In this video, we're going to talk about circuits. Voltage, current, resistance, series, parallel circuits, and things like that. So let's say if you have a battery, and it's connected to a resistor.
Now let's say the voltage of the battery is 12 volts, and we have a 3 ohm resistor. What is the current that flows through the circuit? Now, there's an equation that is associated with Ohm's Law. Voltage is equal to current times resistance. The triangle is basically the change in voltage, or the potential difference, between two points in the circuit.
The voltage across the resistor is 12 volts, is equal to the voltage of the battery. So V is 12. The resistance is 3 ohms. So to solve for the current, which is represented by I, we need to divide both sides by 3. 3 divided by 3 is 1, and 12 divided by 3 is 4. So the current is 4. And the unit for current is amps. So there's 4 amps flowing through the circuit. Now, in which direction is the current flowing in the circuit?
Is it flowing in the clockwise direction? Or counterclockwise? In a direction that is opposite to a clock. Current is going to flow from the positive terminal of the battery to the negative terminal of the battery So it's going to leave the battery from the positive side. It's going to flow through the resistor and It's going to return back to the battery Now there's some things that you need to know What happens if you increase the voltage of a circuit?
So let's say if we increase the voltage of that circuit that we had, what would happen to the current? At constant resistance, as you increase the voltage, the current increases. Voltage and current are directly related.
Now, let's say if we keep the voltage the same. Let's say if we use the same 12 volts, but let's say if we increase the resistance from 3 ohms to 6 ohms. What's going to happen to the current if we increase the resistance?
The more resistance that a circuit has, the less current that will flow through it. So the current will decrease. Current and resistance are inversely related.
Now what exactly is voltage? How would you define voltage? Voltage is the energy per unit charge. So voltage is energy divided by charge.
You can think of voltage as the amount of energy that a charge like an electron has. Electrons, they have negative charges. Protons have positive charges.
So the more energy that an electron has, the greater the voltage. Energy is measured in joules and charge is measured in coulombs. So 1 volt is 1 joule per coulomb. Now in physics, you might have seen this equation.
W is equal to negative Q delta V. W represents work. So work is equal to charge times the change in voltage. Work is measured in joules. Q is measured in coulombs, that's the charge.
And voltage is joules over coulombs. So you can see that the unit coulombs cancel, giving you joules. So V is basically W over Q, if you ignore the negative sign.
But if you want to put it in there, it's like negative Q. So work is associated with energy. Work and energy, they have the same unit. They're measured in joules. Now what about current?
What exactly is current? Current is a rate. Current is Q divided by T.
It's the amount of charge that flows through a current. point per unit time so it's columns over seconds so one amp is equal to one clue per second and Q is simply current times time now the next variable that we need to go over is power what do you think power is power is a rate is the rate of energy being transferred so power is measured in its work divided by time The unit for power is watts. One watt is one joule per second.
So it's how fast energy is being transferred. If you can transfer 100 joules in 2 seconds, Then the power is 50 watts. Now, if it takes you 10 seconds to transfer 100 joules, that means this system is slower in transferring energy. The power is 10 watts.
Now, these two systems, they're doing the same work. transferring the same amount of energy, but the first system is faster. It can transfer 100 joules of energy in two seconds. The second system is slower. It transfers 100 joules, the same amount of work, but in a longer time, 10 seconds.
So power is the rate of energy flow. If you can transfer energy at a fast rate or at a faster rate. the power is greater so power is work divided by time but now how does power relate to voltage current and resistance so if power is work divided by time and we know work is proportional to charge times voltage let's ignore the negative sign then power is charge times voltage divided by time and we said that current is equal to the charge divided by time. The lowercase q and capital Q they both represent charge.
Typically capital Q is usually I normally use it to represent total charge whereas lowercase q might be the charge of a particle. Charge divided by time is the current so power times voltage is equal to the current. This is one equation that you want to keep in mind. Now there's some other equations that you want to know as it relates to power. So we said P is equal to VI and voltage is equal to current times resistance.
So if you replace V with IR, you're going to get this equation P equals IR times I. I times I is I squared. So power is equal to I squared times R. This is another equation that you might find useful. Now starting back with the original equation, P equals VI, let's solve for I in the Ohm's Law equation.
So V divided by R is equal to the current. So this time we're going to replace the current with V over R. So current is V over R, V times V. is V squared so power is also equal to V squared divided by R now let's go back to the circuit that we had in the beginning where we had a 12 volt battery and a 3 ohm resistor and we said the current that flows through the circuit is 4 ohms so here's a question for you How many electrons flow through this circuit in five minutes? How can you calculate that answer?
In five minutes, how many electrons passes through this point? Q is equal to IT. Q represents the total charge that passes through a point.
I is the current, or the amount of charge that passes through it per second. So, if the current is 4 amps, that means... that four columns of charge passes through that point in a single second. So now we need to find the time in seconds. Let's convert five minutes into seconds.
Now the 60 seconds in a minute. So in 5 minutes, there's 300 seconds. So it's going to be 4 coulombs per second, which is 4 amps, times 300 seconds. The unit seconds will...
cancel and we can get the total charge that passes through that point in five minutes so it's four times 300 which is 1200 columns now to find the number of electrons you can convert charge to number of electrons So let me just write this here. Q is 1200 coulombs. Now you need to know the value or the charge of a single electron. One electron has a charge of 1.6 times 10 to the negative 19 coulombs. Now this value is rounded.
It's really like 1.602. So if you start with 1200 clumes over 1, and if you convert it... There's one electron per...
1.6 times 10 to the negative 19 coulombs. Let me get rid of this. And let's put that in the bottom. So that the unit coulombs will cancel.
So the charge divided by 1.6 times 10 to the negative 19. That's equal to the number of electrons that flows through this point. So therefore, 7.5 times 10 to the 21 electrons are flowing at that point in 5 minutes. So in the course of 5 minutes, that's how many electrons has traveled past that point.
Now, you need to be careful. You need to know the difference between conventional current and electron flow. conventional current describes the flow of positive charge so in this circuit the current is flowing clockwise that is not the electron flow the electrons are are flowing counterclockwise.
So anytime the current is flowing in one direction, let's say if it's flowing to the right, the electrons are flowing in the opposite direction. They're flowing to the left. Just keep in mind current, or conventional current, represents the flow of positive charge. So if the positive charges are moving to the right, the electrons will move in the opposite direction.
They're going to move to the left. So when you're dealing with circuits, and when you hear the word current, Typically, they're describing conventional current. Now, let's talk about power.
What is the power delivered by the battery, and what is the power absorbed by the resistor? In this circuit, the battery provides energy to the circuit, and the resistor absorbs that electrical energy. and converts it to heat energy.
So what is the power supplied by the battery to the circuit? So power is equal to voltage times current. So the voltage is 12. And the current that flows from the battery is 4 amps.
So 12 times 4 is 48. So the battery is delivering 48 watts to the circuit. Now what about the resistor? How much power does the resistor absorb? Well we know if the battery provides 48 watts, the resistor must consume 48 watts.
It must absorb that 48 watts and transfer it to heat energy. Now this is assuming that there's no loss of energy in the wires, but the wires do contain a small amount of resistance, so a small amount of energy loss does occur through the wires. But let's say if the wires are perfect, and only the battery provides the energy and the resistor absorbs the energy. In such a situation, we can calculate the power absorbed by the resistor. let's use this form of the power equation P equals I squared R which we divide earlier in this video.
The current that flows through the resistor is 4 amps and the resistor has a value of 3 ohms. 4 squared is 16, 16 times 3 is 48. So notice that the circuit is balanced. The battery delivers 48 watts and the resistor absorbs 48 watts. Everything must be equal. Now here's a question for you.
We said that power represents the rate of energy or the the rate of energy flow. How many joules of energy are being transferred by the battery in five minutes? So how much energy does the battery loses in five minutes?
Now the power is 48 watts. You need to know what that means. We said that one watt represent a transfer of one joule in one second. So in one second It's the If the power is 48 watts, the battery is transferring 48 joules every second.
So if it delivers 48 joules in one second, that means that it's going to deliver 480 joules in 10 seconds. And it's going to deliver 4800 joules in 100 seconds. So how much energy will it deliver in 5 minutes?
Now we said that power is equal to work divided by time. If you rearrange the equation, you'll see that work, which represents the energy transferred or the energy delivered, work is power multiplied by time. So the power is 48 watts and the time is 300 seconds, which equates to 5 minutes. So 48 times 300 is 14. thousand four hundred so in five minutes the battery delivers fourteen thousand four hundred joules of energy if it delivers that much energy that means the battery has lost fourteen thousand four hundred joules of energy and resistor transferred that energy to heat energy so the battery lost energy and that energy eventually escaped as heat Now you can also solve it using a conversion. It helps to understand how the units work along with the equation.
So let's say if we started with 48. So instead of writing 48 watts, we can write 48 joules per second. per second actually let's use that later let's start with five minutes let's start with the time so we have a time of five minutes and we can convert it to seconds one minute is equal to 60 seconds Now, instead of multiplying by 48 watts, let's replace watts with what it equates to, that is joules per second. So 48 watts means that 48 joules are being transferred every second. So the unit minutes cancel and the unit seconds cancel, and so now you have the energy in joules.
45 times 60, which is 300, times 48 will give you the same answer of 14,400 joules. transferred. Now consider the following two circuits. Which one is a series circuit and which one is a parallel circuit? And also, why?
Why did you choose the answer that you chose? It turns out that the series circuit is the one on the left, and the one on the right is the parallel circuit. A series circuit is a circuit in which current only flows in one direction. So in this circuit, the current has no choice but to flow through each resistor.
It can only flow in that single path. The series circuit is only one path for the current to flow. And the current that flows to two resistors that are in series is the same current. So if you have a 4 amp current flowing through the first resistor, four amps will flow through the second resistor. So whenever you have two resistors in series, they will have the same current, because there's only one path for that current to flow.
Now, in a parallel circuit, there's multiple paths for the current to flow. The current could flow through one path, or it can flow through another path. Now, the current that flows through those two resistors, they don't have to be the same.
Unless the resistors have the same value, that's the only case where the currents will be the same. Other than that, if the resistors are different, then the current will be different in a parallel circuit. But in a series circuit, it doesn't matter the value of the resistors. If this was a 5 ohm resistor and this was a 10, the current that flows through them will be the same.
So in a series circuit, you have the same current that flows through the resistors that are in series with each other. In a parallel circuit, which contains multiple paths for the current flow you have the same voltage across the resistors so let's say if we have a 10 volt battery the voltage across the first resistor will be 10 volts and the voltage across the second resistor will be 10 volts because the two resistors are attached to the same two wires so in a parallel circuit whenever two resistors are parallel to each other they're gonna have the same voltage across them, and whenever two resistors are are in series with each other, the same current will flow through them. So make sure you're aware of those fundamental concepts because it's going to help you to solve complicated circuits, which we'll go over later in this video.
Now before we work on some calculations with series and parallel circuits, let's focus on Ohm's Law and how voltage and current applies to a resistor. So let's say if you have a resistor. and there's a voltage of 4 volts at these two points. What is the current that flows to the resistor?
And let's say the resistance is 2 ohms. Now you might be thinking, V equals IR. Now, the current is not 2 amps. You shouldn't plug in 4 into the voltage. What this equation really means, it means that the change in voltage is equal to the current times the resistance.
Voltage is described as electric potential or sometimes potential difference. The potential difference between these two points, the change in voltage, is zero. So therefore, no current flows through this resistor. In order for current to flow through resistor, you have to have a difference in voltage. If the voltage is the same, the current will not flow through it.
A good way to think about it, imagine if you have a pipe with water. If the pipe is leveled, No water flows. The water is going to remain stagnant. It's just going to sit there. But now, let's say if you raise one side of the pipe relative to the other.
Well, water is going to flow from a high position to a low position. to a low position, it's going to flow downward. Because at position A, position A has more gravitational potential energy because it's at a higher height than position B.
Gravitational potential energy... energy is mgh. So the higher an object is above ground, the more potential energy that it has.
As it falls, that potential energy is converted to kinetic energy and the object speeds up. It gains speed. Voltage works the same way.
Voltage is known as electric potential. It's very similar to potential energy. Think of... So this situation here relates to this situation.
What we have is a level pipe and no water is going to flow because the voltage is the same. Now let's say if the pipe wasn't leveled. Let's say if this is 12 volts at this point, and this point is 4 volts. And we have a 2 ohm resistor.
So in this case... We have like a pipe that has water, and let's say this is ground level. At point A, it's 12 meters above the ground.
And at point B, it's 4 meters above the ground. So therefore, the height difference is 8. It's really the height difference that's important, not really how high it is above the ground. So likewise, in this circuit...
is the voltage difference between these two points that's important. So using Ohm's Law, the change in voltage, or the potential difference between the two points across the resistor, is 8 volts, that's equal to I times... times the resistance of 2 ohms.
So 8 divided by 2 will give us a current of 4 amps. So 4 amps will flow through the resistor. Now notice that current, it flows from high voltage to low voltage. And it makes sense because water, it flows from a high position to a low position. It's not going to flow from low to high, it flows from high to low.
so likewise current electric current it flows from a high voltage to a low voltage now if you think about it from a fundamental viewpoint the side of the resistor that has a higher voltage is positive and the side with the lower voltage is negative now we said that current represents the flow of positive charge so let's say if this side is positive and this side is negative and if you put a proton in the middle. In which direction will the proton accelerate towards? Will it accelerate towards the positive plate or towards the negative plate?
We know that like charges repel and opposite charges attract. So therefore, the proton is repelled by the positively charged plate. So it wants to move away from the left side. It doesn't want to be next to the the positively charged plate. However, the proton is attractive.
to the negatively charged plate so it's going to accelerate towards the negatively charged plate since it's attracted to it therefore a positively charged particle it's going to move away from the side with the higher voltage which is positive and it's going to flow towards the side with the negative voltage or the side that has less voltage because it's attracted to it attracted to the negative charge now for a lecture the situation is different If you place an electron between these two plates, the electron will be repelled by the negatively charged plate, and it will want to go towards the positive plate. Likewise, an electron in its resistor. It's going to be attracted to the positive plate. the positive charge and it's repelled by the negative charge so it's going to flow in the opposite direction so current conventional current which represents the flow of positive charge flows from a high voltage to a low voltage electrons will flow in the opposite direction electrons will flow from low voltage to high voltage it's going to flow towards the positive charge so make sure you understand those fundamental concepts so now let's work Let's work on some example problems.
So let's say if we have a resistor, and this side is at 16 volts, and this side is at 6 volts, and the value of the resistance is 5 ohms. What is the magnitude of the current, and what is the direction of the current through this resistor? So first, what is the change in voltage?
The change in voltage is 16 minus 6. So we have 10 volts across the 5 ohm resistor. So delta V is 10. The resistance is 5. 5 10 divided by 5 is 2 so the current that flows through the resistor is 2 amps now conventional current it flows from high voltage to low voltage so in this example it's going to flow from right to left so it's going to go to the side with the lower voltage Now what about this example? Let's say if this is at negative 12 and this is at negative 30 and we have a 6 ohm resistor. Calculate the amount of current that flows through the resistor and also determine the direction. So what is the difference in voltage?
What is the difference between negative 12 and negative 30? So, negative 12 is greater than negative 30. On a number line, this is 0, here's negative 12. and here's negative 30 as you travel to the right on a number line the value increases so negative 12 is larger than negative 30 so this is the positive side and this is the negative side therefore we know that current is going to flow from high voltage to low voltage so in this case is going from left to right Now, negative 30 and negative 12, they differ by 18 volts. If you take the larger value and subtract it by the smaller value, you're going to get 18. Whenever you have two negative signs next to each other, it's going to change to a positive sign.
So this is negative 12 plus 30, which is 18. So using Ohm's Law, V equals IR, the change in voltage is 18, the resistance is 6, 18 divided by 6 is 3. So we have a current of 3 amps flowing to the right. Try this example. Let's say if there's 20 volts at the bottom, and at the top, it's negative 8 volts.
and we have a 7 ohm resistor. Calculate the current and the direction of the current. So let's find the potential difference.
The change in voltage, or potential difference, is... 20 minus negative 8 which is the same as 20 plus 8 so these two they differ by 28 volts so 28 divided by 7 is 4 so the current is 4 amps and 20 volts is higher in potential than negative 8 so the current flows from high to low so in this case it's going up Now let's say if you have a 5 ohm resistor and a current of 4 amps flowing through it. And let's say at this point the voltage is 30 volts.
What is the voltage at this point? Feel free to pause the video and work out this example. So let's find the voltage difference across the resistor. So the voltage across the resistor, using Ohm's Law, is the current that flows through the resistor, which is 4 amps, times the resistance itself.
So the potential difference is 20 volts. So if you take an ammeter... And you connect it across the resistor.
It's going to read 20 volts. So now, what is the voltage at the other side? We know it's different by 20 from 30. 30. So, is it 20 volts higher or 20 volts lower?
Is it 30 plus 20, which will give us 50, or is it 30 minus 20, which will give us 10? Now, the key to figuring out the answer is the direction of the current. Current flows from high voltage to low voltage. So, the 30 volts has to be the higher voltage.
The other side is the lower voltage, so we got to subtract instead of adding. So 30 minus 20. is 10 so it's 10 volts on this side now let's try another example so let's say if the voltage at this point is 8 volts And there's a current of 5 amps flowing through a 7 ohm resistor. What is the voltage at that point? So let's find the voltage across the resistor. So what is 5 times 7?
V equals IR. A current of 5 amps times a resistance of 7 will give us a voltage difference of 35. So since current is flowing from the bottom to the top, the voltage at, let's say, point B has to be larger than point A. So it's 8 plus 35 instead of 8 minus 35. 8 plus 35 is 43. So this is the voltage at this point.
Now let's try another example. Let's say if we have a current of 6 amps that passes through an 8 ohm resistor. And let's say that the voltage at point B...
negative 50 volts what is the voltage at point A so at point A is supposed to be higher than it is at point B since current flows from high to low Now what is the voltage across the resistor? So we have 6 amps of current flowing through an 8 ohm resistor. V equals IR, 8 times 6 is 48. So the voltage across the resistor is 48 volts.
So now should we add 48 to negative 50 or should we subtract 48 from negative 50? Since A is at a higher potential, we need to add. plus 48 is negative 2. Negative 2 is higher than negative 50 on the number line. So here's the last example with these questions, but I really want you to understand this concept. Now, let's say if the voltage here is at negative 40, and we have a current of 9 amps that passes through an 8-ohm resistor.
What is the voltage at the bottom? so the voltage across this resistor is nine times 8 which is 72 volts so the current is flowing from high to low so at point B, point B is lower than point A so we need to subtract 72 from negative 40 to get the potential at point B since we're looking for the lower voltage so negative 40 minus 72 is negative 1, 12. I believe that's right. And so, that's it. That's how you can find the potential.
So, if you're looking for the higher potential, make sure you add the potential difference to it. If you're looking for the lower potential, subtract. Now let's work on a series circuit that contains two resistors.
Now let's say the first resistor has a value of 7 ohms, and the second one is 3 ohms. And the battery has a voltage of 20. So this is the positive side, and this is the negative side. Now, what is the current that flows through the series circuit? In order to find the current, you need to find the total or the equivalent resistance. So let's call this R1 and R2.
For a series circuit, the total resistance is simply the sum of the individual resistors. So 7 plus 3 is 10. So the total resistance is 10 ohms. So you can combine these two resistors as if it was a single resistor. So if you want to draw an equivalent circuit, it would look like this so we have a 20 volt battery and a 10 ohm resistor So now what is the current that flows through this equivalent 10 ohm resistor? So using V equals IR, the voltage across the 10 ohm resistor is 20 volts, and the resistance is 10. 20 divided by 10 is 2, so we have 2 amps flowing through the 10 ohm resistor.
Therefore, in our original circuit, we have 2 amps of current. Now, let's focus on the first resistor. We know that the current that flows through R1 is 2 and it's the same as the current that flows through R2. Whenever you have two resistors in series, the current that flows through them is going be the same.
So, if we have 2 amps of current flowing through a 7 ohm resistor, what is the voltage across the resistor? Once again, you can use V equals IR, just in a different way. So we have We have 2 amps of current flowing through a 7 ohm resistor, therefore the voltage across that resistor is 14 volts.
Now what is the voltage across the 3 ohm resistor? So using the same equation V equals IR, we have 2 amps of current flowing through a 3 ohm resistor, so the voltage is 6 volts. Notice that these two voltages is equivalent to this voltage.
This is related to Kirchhoff's or Kirchhoff's voltage law, which states that the sum of the voltage around a closed circuit adds up to zero. Now you might be wondering why does it add up to zero? Because if you add 20 plus 14 plus 6 it appears to be 40. but it's not one of the values is negative and the others are positive remember the battery delivers energy and resistors absorb of energy. Now I've seen different sign conventions when you're deciding which ones are positive, which ones are negative, but regardless of which way you do it, the sum will always be zero. Now let's say at this point the voltage is zero.
That means at the top the voltage must be 20 since across the battery the potential difference is 20. The positive side is the side of the higher voltage. Now let's analyze it going clockwise. So as we travel this way, some textbooks will say that it's negative 20 volts.
If you go in that direction, others will say it's positive 20. You may have to simply pick a method that helps you to understand it. One way I like to see it is I like to focus on the circuit as a system. Elements that increase the energy of the system, I like to view it as a positive contribution. And other elements that decrease the energy of the system, a negative contribution.
So this battery, it delivers power to the system. It increases the voltage of the system by 20. So I'm going to say positive 20. And as we travel in that direction, we're going from 0 to 20. Now as we travel through resistor 1, this voltage here is 6 volts. Notice that the potential difference is 6, which is the difference between 0 and 6. So as we go from 20 to 6, we've experienced a voltage drop.
And it makes sense. Resistors, they consume energy. And since voltage represents the electric potential energy per unit charge, they're going to drop the voltage, they're going to consume it, they're going to decrease it.
So, the voltage across the resistor is negative 14. It decreases the energy of the circuit by absorbing energy from that circuit. And now, as we travel from 6 to 0, the second resistor is going to drop the voltage by 6 volts. So the total voltage around the circuit is zero, which agrees with Kirchhoff's voltage law.
Now let's say if we travel in the other direction. It's still going to add up to zero volts. So if we travel in this direction, we went from zero to six, so the voltage increased by six.
And through resistor 1, we went from 6 volts to 20 volts. So it increased by 14 volts. Now, going through the battery, we went from 20 to 0. So the voltage decreased by 20. 6 plus 14 minus 20 is still 0 volts.
So it really doesn't matter what direction you travel. It's still going to add up to zero volts if you look at the potentials, the absolute potentials across the circuit. If you view it from that way, it's going to be a lot easier to write the Kirchhoff voltage expression. Now, let's calculate the power delivered by the battery. So we know that power is voltage times current.
At least that's one form of the equation that you can use. So the voltage of the battery is 20 volts, and it dishes out 2 amps of current. 20 times 2 is 40. So the 20 volt battery releases 40 watts of energy, or 40 watts of power, I should say, which is 40 joules per second. Now what about the 7 ohm resistor?
How much power does it absorb? So let's use the second form of the power equation, P is equal to I squared times R. So the current that flows through the R1 resistor is 2 amps.
So I is 2, and let's not forget to square it. And the resistance of R1 is 7. 2 squared is 4, 4 times 7 is 28. So R1 absorbs. 28 watts of power.
It absorbs 28 joules in a single second. Now what about the power absorbed by R2? Let's use the third form of the equation. Power is equal to V squared divided by R. The voltage across the 3 ohm resistor is 6 volts.
volts, so 6 squared, divided by the value of the resistance, which is 3. 6 squared is 36, 36 divided by 3 is 12. So R2 absorbs 12 watts of power, or 12 joules every second. So notice that the power delivered by the battery is equal to the total power absorbed by the two resistors. The energy must be conserved in this circuit.
The amount of energy that the battery puts into the circuit should equal the amount of energy that the resistors absorb from the circuit. Now let's try another problem. So this time, we're going to have three resistors.
So let's say that this one is 3 ohms, this is 4, and this is 5 ohms. And the voltage of the battery is 48 volts. Calculate the current in this circuit. Now, in this particular circuit, you can redraw it like this. It's the same as viewing it this way.
Let's say R1 is the 3-ohm resistor, R2 is the 4-ohm resistor, and R3 is the 5-ohm resistor. So what is the current that flows through the circuit? So we still have a series circuit because...
The same current that flows through R1 is the same current that flows through R2 and R3. So the total resistance in a series circuit is the sum of the resistors connected in series. So it's 3 plus 4 plus 5. five so the total resistance is 12 in the series circuit that's the first thing you want to do you want to find the total resistance next you want to find the current the total voltage or the voltage of the battery is equal to the total current times the total resistance the voltage of the battery is 48 and the total resistance is 12 so 48 divided by 12 is 4 so we have 4 amps of current. flowing through the circuit. Now calculate the voltage drop across each resistor.
So to find the voltage drop is simply V equals IR. So it's the current times the resistance. So for the first resistor, we have a current of 4 amps flowing through a 3 ohm resistor.
4 times 3 is 12. so we have a voltage drop of 12 across that resistor now what about the voltage drop across the four-ohm resistor so we have four amps of current flowing through a four-ohm resistor V equals IR, 4 times 4 is 16. And for the 5 ohm resistor, 4 times 5 is 20. Notice that if you add 12 plus 16 plus 20, you get a total of 48, which satisfies Kirchhoff's voltage law requirement. So now that we have the current and the voltage across every resistor, let's find the absolute voltages at each point along the circuit. So let's assume that at this point the voltage is zero. What is the voltage at this point?
Since the difference is 20 and the current is flowing in a downward direction, Point A has to be higher than point B. Let's call this point C and point D. So that would mean that point A is at 20 volts.
Now, the voltage difference here is 16, so that means this point is at 36 volts. And over here, it has to be 48. It's 12 higher than this value. Now, let's say... If the voltage at this point was 30 volts, what is the voltage at this point?
So we know it's going to be 20 volts higher, since the current is flowing in a downward direction. So it's going to be 30 plus 20, which is... is 50 and at this point it's going to be 16 volts higher than 50 so it's going to be 56 volts I mean not 56 but 66 volts and at this point it's 12 volts higher which is 78 so the difference between these two is still 48 So that's how you can calculate the absolute potential around a circuit. But you need to know the voltage of one point around it.
So if we use Kirchhoff's voltage law, if we try to prove it, going in this direction, we're going from 30 to 78 so the battery is increasing the voltage of the circuit by 48 and if we travel through the first reason sister we're going from 78 to 66 so the voltage drops by 12 volts so negative 12 and as we travel from 66 to 50 the voltage drops by 16 volts and as we travel from 50 to 30 the voltage drops by 20 volts so 48 minus 12 minus 16 minus 20 adds up to zero Now let's work on a parallel circuit. This is a circuit with multiple paths for the current to flow. Now let's say if we have a 12 volt battery, a 3 ohm resistor, and a 4 ohm resistor. Now let's assign a voltage of 0 volt at this point. So anywhere along this line, the voltage is 0. Now at the top, the voltage is 12. It's 12 at this point.
point and at that point. So the voltage across the 3 ohm resistor and the 4 ohm resistor is 12. Now keep this in mind, the current that passes through two resistors in series is the same current. the voltage across two resistors that are parallel to each other is the same voltage.
How can we find the current that flows through the 3-ohm resistor? So we can use Ohm's Law. Voltage is equal to current times resistance. So the voltage across the 3-ohm resistor is 12, and the resistance is 3. So 12 divided by 3 is 4. Therefore, 4 amps of current flows through the 3-ohm resistor. Now what about the 4-ohm resistor?
The voltage across the 4-ohm resistor is still 12 volts. This is 12, and this point is 0. But the resistance is 4. 12 divided by 4 is 3. So 3 amps flows through the 4-ohm resistor. So how much current leaves the atmosphere?
the battery the amount of current that leaves the battery is the total of these two currents so seven amps leaves the battery now that Number, the 7 amps that we got, is related to Kirchhoff's Current Law. Now let's focus on this point right here. Kirchhoff's Current Law states that the current that enters certain point equals the current that leaves that point so seven amps of current enters this point four amps travels downward and the other three amps go in this direction.
So as you can see, the current that enters the point equals the current that leaves the point. And that's the main idea behind Kirchhoff's current law. Now, what is the equivalent resistance of the 3-ohm and the 4-ohm resistor?
There's two ways you can find the equivalent resistance. You can use the equation, or you can use Ohm's Law. V equals IR. If you know the voltage of the battery, which is 12, and the total current of the circuit, which is 7, you can calculate the equivalent resistance.
So you can view this entire circuit as one single resistor and the seven amps flows through that single equivalent resistor so that's why we can use V equals IR in that way so the equivalent resistance is simply 12 divided by 7 it's 12 volts over 7 amps now there's another way in which you can calculate that same answer and it's used in this equation for parallel circuits 1 over R equivalent resistance or total resistance is equal to 1 over R1 plus 1 over R2. And if there's a third resistor plus 1 over R3. So R1, let's say it's 3 and R2 is 4. So let's solve for Req. So to do that we need to multiply both sides by 12. We might need an R as well.
Let's multiply both sides by 12R. So 12R times 1 over R is 12. 12R times 1 third, that's 12R. That's going to be 4r. 12 divided by 3 is 4. And 12r times 1 fourth is 3r.
4r plus 3r is 7r. And if you divide both sides by 7, you'll get the same answer. The equivalent resistance is 12 over 7, which is 1 point something.
Now let's calculate the power delivered by the battery and the power absorbed by the two resistors. So the power delivered by the battery is P is equal to VI. The voltage of the battery is 12 and the current that it delivers is 7. 12 times 7 is 84. So the battery delivers 84 watts of power.
Now what about the 3 ohm resistor? How much power does it absorb? So let's use I squared R. So the current that flows through the 3 ohm resistor is 4 amps, and its resistance is 3. 4 squared is 16. 16 times 3 is 48. So the 3 ohm resistor absorbs... 48 jewels of energy every second or 48 watts now it's calculate the power absorbed by the other resist using V squared over R so the voltage across this 4 ohm resistor resistor is 12 volts, so it's going to be 12 squared, divided by the resistance of 4. So, 12 squared is 144, but what's 144 divided by 4?
Now, another way you can get the answer, instead of doing 12 squared, you can divide 12 by 4. 12 squared is the same as 12 times 12. 12 divided by 4 is 3, and 3 times 12 is 36. So therefore, 142 is 3. 44 over force 36 if you divide first before you use the exponent it's easier to get the answer so the four ohm resistor absorbs 36 watts of power 36 plus 48 is 84 so it makes sense the battery delivers 84 watts of power and the two resistors absorb a total of 84 watts of power so therefore Therefore energy is conserved for this problem. Now let's try another example of a parallel circuit. This time we're going to have three resistors instead of two. So we're going to use a 24 volt battery.
The first resistor is going to have a value of 3 ohms. The second is going to be 4 and the third is going to have a value of 6 ohms. Calculate the current that flows through each resistor. So we know the voltage across the three resistors connected in parallel is the same.
They're all connected to the same battery, so the voltage across each resistor... is 24 volts so V equals IR to find the current that flows through each respective resistor we need to take the voltage across that resistor and divided by the resistance to get the current V over R equals I so 24 divided by 3 will give us a current of 8 amps that flows through the first resistor 24 divided by 4 will give us a current of 6 amps flowing through the second resistor. And 24 divided by 6 will give us a current of 4 amps flown through the third resistor.
So therefore, what is the current that leaves the battery? The current that leaves the battery is the sum of the three currents. It's 4 plus 6 plus 8. So 18 amps leaves the battery.
Now at this point, here's a question for you. how much current flows along this part of the wire. Now, if you think about it, 18 amps enters the point.
8 amps goes down. 18 minus 8 is 10, so 10 amps must be flowing to the right. Now, as we approach the second point, let's get rid of the 24. We know that 10 amps enters the second point. 6 amps travels in a downward direction. 10 minus 6 is 4, so 4 amps flows this way.
And 4 amps flows through this wire. Now at this point, the 6 amp current and the 4 amp current, they join forces and they add to form like a bigger river. And so this is going to be 10 amps.
Then the 8 amp... current and the 10 amps they will combine at this point and the current will be 18 amps and then 18 amps will flow through the battery and then the cycle will be repeated so as you can see you can apply Kurt coughs current law at these different points So let's work on some questions dealing with Kirchhoff's current law. So let's say if we have a junction and 10 amps of current flows into the point and 3 amps travels this way.
What is the current that leaves that point? to the right. So if 10 amps enters the point, 10 amps must leave the point. So 10 minus 3 is 7, so 7 amps travel in that direction.
Now let's change it. So let's say if we have the same picture and 10 amps is flown in this direction and three amps is flown in this direction what's the current that is traveling in this direction so notice the amount of current that travels towards the point so the 10 amps it flows towards the point so it's going into it the 3 amp current also goes towards the point so we have 13 amps traveling towards the point therefore 13 amps must leave away from the point. Now what if we have a more complicated circuit? Let's say if 5 amps of current travels towards this point and then 8 amps of current goes in this direction and 9 amps travels this way what is the current that travels to the right so one way to think about it is the amount of current that is going towards the point and the amount that is leaving the point it's the same as the amount that enters the point and the amount that leaves it. So the amount that goes towards the point is 5 plus 8 plus 9. These three currents, they're traveling towards the point.
And if you add them, you should get 22 amps. So therefore, the amount of current that moves away from the point must be 22 amps. Now, let's say if you have a current of 8 amps going in that direction, and 5 amps going this way, and let's say 4 amps is flowing in this direction.
What is the missing current? How would you figure it out? So how much current is going towards the point? The 4 amps and the 8 amp current is going towards the point. So we have a total of 12 amps going towards that point.
Now how much current is leaving the point? total must be 12 amps as well if 12 amps moves towards the point 12 amps must move away from the point so the 5 amp current is moving away so So 5 plus what number is 12? The answer is 7. 5 plus 7 is 12. So we have 12 amps moving away from the point and 12 amps moving towards the point. Now let's try a more complicated example. So let's say if we have 12 amps of current traveling in that direction, 6 amps of current going this way.
5 amps go in that direction, and 4 amps travel in this direction. What is the... let's see.
What is the current that leaves in that direction? What is the answer? So, the current that enters or moves towards the center, particularly this point, is equal to the current that leaves the point.
So, which currents are going to be the current that enters or moves towards the center? towards the point the 12 amp current is going towards it the 4 amps enters the point and the 5 amps as well so on the left side we have 12 plus 4 plus 5 Now what's the current that leaves the point? So we have the 6 amp current, it leaves it, and the missing current, which we'll call I. So I plus 6, it leaves the center point.
so 12 plus 4 is 16 16 plus 5 is 21 and 21 minus 6 is 15 so therefore that's the missing current 15 amps leaves the junction so as we can see 15 and 6 is 21 and 12, 4, and 5 also adds to 21 Now let's say if we have a combination of a series and a parallel circuit. Now let's say if we have a 6 ohm resistor and a 3 ohm resistor. Actually, let me change numbers.
Let's make this a 12-ohm resistor and a 4-ohm resistor. And let's say that this one is a 7-ohm resistor. And the voltage of the battery is 50 volts.
What would you do to solve the circuit? How can we find the current that flows through every resistor? And how can we find the voltage across each resistor? With the information that we're given.
Now, if all you have is just the resistors and the voltage, it might be wise to calculate the equivalent resistance. So notice that the 4 ohm and the 12 ohm resistors are connected in parallel. that means that the voltage across them is the same so to find the equivalent resistance between the forum and the 12 ohm resistor we can use this equation req or one of our IQ is 1 over r1 plus 1 of our 2 so solving for our EQ it's 1 over R 1 plus 1 over R 2 raised to the minus 1 power so if you type this in the calculator 1 over 4 plus 1 over 12 raised to the negative 1 you should get 3 So therefore, these two resistors, you can represent them as a single resistor of 3 ohms.
So if we redrew the circuit, it's going to look like this. So here is the... the 7 ohm resistor, here is the 50 volt battery, and here is the equivalent 3 ohm resistor, which is the combination of those two resistors. Whenever you add two resistors in series, the total resistance increases because less current will travel through them. It's harder to travel through two resistors instead of one.
Now whenever you add two resistors in parallel, the current increases so therefore the total resistance goes down. The reason why the current increases whenever two resistors are connected in parallel is because you provide another pathway for electrons to flow. it's easier to travel through a two-lane highway than a one-lane highway. There's more space.
So that's why resistors in parallel decreases the total resistance, and resistors in series increases the total resistance. Now, let's go ahead and calculate the total resistance of the circuit on the right side. So, the 7-ohm resistor is in series with the 3-ohm resistor. So the total resistance.
is 10 so using V equals IR we have a voltage of 50 a total resistance of 10 therefore the current that flows through this circuit is 5 amps so we have 5 amps that leaves the battery so we can get rid of this circuit at this point So we know that 5 amps of current flows through the 7 ohm resistor. So what is the voltage across the 7 ohm resistor? V is equal to IR.
So the current is 5, the resistance is 7. 5 times 7 is 35. Now what is the voltage across the 12-ohm resistor? Now based on Kirchhoff's voltage law, we know it has to be 15. And the reason for that is because 15 and 35 adds up to 50. The voltages across the resistors have to add up to the voltage of the battery. Even though one of them might be negative and the other might be positive, they should still equal to each other. so we know the voltage across the 4 ohm resistor and a 12 ohm resistor is 15 volts since they're connected in parallel to each other so now let's calculate the current that flows through the 4 ohm resistor and the 12 ohm resistor so the voltage across the 4 ohm resistors 15 divided by the resistance of 4 so 15 divided by 4 is about 3.75 so 3.75 amp of current flows through the 4 ohm resistor and 15 divided by 12 is 1.25 so 1.25 amps flows through the 12 ohm resistor notice that these two currents the currents that leave this point is equal to the current that enters the point 3.75 plus 1.25 adds up to 5 in harmony with Kirchhoff's current law Now to make sure that everything is correct, the power...
must be balanced. The power delivered by the battery must equal the power absorbed by the resistors. So for the battery, it has a voltage of 50, a current of 5 amps, so P equals VI.
50 times 5 is 250. So the battery delivers 250 watts of power. Now what about the 7 ohm resistor? How much power does it absorb?
So using I squared R, We have a current of 5 amps. so I squared is 5 squared which is 25 and 25 times 7 is 175 if you have seven quarters that's a buck seventy-five so the power absorbed by that resistor is 175 watts now what's the power absorbed by the 4 ohm resistor now let's use V squared over R the voltage across it is 15 15 squared is 225 if you divide up by 4 you'll see that the power absorbed by the 4 ohm resistor is 56.25 watts. Now what about the 12 ohm resistor?
Using V squared over R, 15 squared divided by 12 that's 18.75 watts. So if we add up the powers absorbed by the three resistors So it's going to be 175 plus 56.25 plus 18. and that indeed adds to 250. So the energy is conserved in this particular example. Now let's say if the voltage at this point is 60 volts. And let's say that we have... a 5 ohm resistor and there's a current of...
I need to choose the right current. Let's say there's a current of 6 amps that flows through this resistor. And let's say this particular resistor has a value of 3 ohms and this one is 6 ohms. What is the current that flows through the 3 ohm resistor? And what is the current that flows through the 6 ohm resistor?
And also, what is the voltage at point A and point B? So without a battery, how... How would you solve this particular circuit?
Feel free to pause the video and work out this example. So how would you figure out how much current is going to flow through the 3 ohm resistor and a 6 on reserve. resistor well let me ask you this which resistor will contain more current is it the 3 ohm or the 6 ohm resistor now we said early in this video that as the resistance goes up the current goes down provided that the voltage is the same and the 3 ohm resistor is in parallel with the 6 ohm resistor which means that the voltage across them is the same.
So the resistor that has the higher value has less current. So the 6 ohm resistor, since it has twice the value of the 3 ohm resistor, it's going to have half the current compared to the 3 ohm resistor. So the 3 ohm resistor is going to have two times as much current than the 6 ohm resistor.
So here's a question for you. What two numbers add up to 6 where one number is twice the value of the other number? we know that 2 plus 4 equal to 6 and 2 times 4 I mean 2 times 2 is 4 so the currents that flow through the two resistors are 2 and 4 so we have 2 amps flowing through the 6 ohm resistor and 4 amps flowing through the 3 ohm resistor and if you calculate the voltage across each resistor V equals IR 3 times 4 is 12 and 6 times 2 is 12 Therefore, the voltage across them remains to be 12. Now, there's another way in which you can calculate the current that flows through each resistor if you have the current that enters that system.
And here's the equation that you can use. Let's call this current I1 and this one I2. So, therefore, this is going to be R1 and this is R2. So let's start with I1.
To find I1, it's going to be the other resistor, R2, divided by the total resistance times the total current that enters the branch, which is IT. IT is 6. So if you plug it in... R2 is 6. The total of R1 and R2 is 3 plus 6, which is 9, times the current that enters the branch, which is 6. 6 times 6 is 36. 36 divided by 9 will give us... the 4 amp current, which is I1. Now, to find the other one, I2, we can use a similar equation.
So, I2 is going to be equal to the other resistor, R1, divided by R1 plus R2, times the total current. So, if this is a 2, this has to be 1. R1 has a value of 3. Definitely. The total resistance R1 plus R2 is still 3 plus 6, which is 9, and the total current is 6. 3 times 6 is 18. 18 divided by 9 is 2, so you get the 2-amp current. So that's the equation that you can use if you want to find a current that flows through two resistors, if you know the resistance, and if you know the current that enters the system, or the parallel circuit system. Now, how can we find the voltage at point A?
So, what we need to do is find the voltage across the 5-ohm resistor. So, V equals IR. The current that flows through the 5-ohm resistor is 6 amps. times the R value of 5. That gives us 30 volts. Now, the voltage at point A, is it greater than or less than 60?
Now, we know that current, it flows in a direction from high voltage to low voltage. voltage to low voltage since it's flowing towards a it's going to flow to the voltage that has it's going to flow to the side with the lower voltage so a is less than 60 so then instead of doing 60 plus 30 it's going to be 60 minus 30 so the potential at a is 30 volts now the voltage across the 6 ohm resistor is 6 times 2 which is 12 so to find the voltage at B we know it's less than that of a since the current is flown towards B so we need to subtract 30 minus 12 is 18 so that's the voltage potential at point B now let's talk about the voltage divider circuit we already drew the circuit it's basically a series circuit with two resistors Now, let's say the voltage of the battery is 100 volts, and this resistor has a value of 100 ohms. And we want to calculate the voltage across this resistor. Now, by adjusting the value of R, you can control the voltage that will be across that resistor.
If R is equal to 100 ohms, if these two resistors are exactly the same, then the voltage across R is going to be half of what the battery is. so it's going to be 50. Now if R exceeds or if R is greater than 100 then the voltage across R will be greater than 50. The maximum voltage that you can get above R or across R is close to 100, you can't exceed this value and the lowest is 0. So let's say if we want a voltage of 90 volts across this resistor, what value should we choose for R? So let's call this R1 and R2. And let's call this the output voltage.
The output voltage is equal to R2 divided by the total resistance times the input voltage, which is the voltage of the battery. So now let's solve for R2. So the output voltage is 90. We're looking for R2. R1 is 100 plus R2.
And the input voltage is 100. If we divide both sides by 100, 90 divided by 100 is 0.9. And so R2 divided by 100 plus R2 is 0.9. So we can put 0.9 over 1 and cross multiply. So 1 times R2 is R2. And 0.9 times 100 plus R2, that's going to be 90 plus 0.9 R2.
And I'm running out of space. So we have R2 is equal to 90 plus 0.9 R2. So let's subtract both sides by 0.9 R2.
So 1 R2 minus 0.9 R2 is 0.1 R2. So 90... divided by 0.1, R2 is about 900 ohms.
So in this case, we can see that if you increase R2, the output voltage across R2 will increase. So if R2 is 900, the voltage across it is 90. The voltage across the 100 ohm resistor must be 10 volts. 10 and 90 adds up to 100. Now, we could see why the equation works mathematically, but let's see if we could understand it conceptually. Why is it that when we increase the value of R2, the voltage across it increases? How can we make sense of it?
So, if you connect a voltmeter... across R2. If the value of R2 goes up, the voltage across R2 goes up. If you increase the value of R2, electrons will prefer to travel through the meter instead of R2. So the current will want to go this way, as opposed to going this way.
Let's say if you're you're driving on the highway or on a road and you have a choice in making a left turn or right turn and let's say if you make a left you can see the traffic the five o'clock traffic jam bump into bumper and it's terrible but if you make a right you can see that there's no traffic there's not many cars in the road and it's just easier to travel on that road which path will you take the path of least resistance or the one with most resistance typically you'll probably drive on the road with less traffic I mean who wants to be in traffic so likewise the electrons they prefer to travel the path of life. least resistance so as you increase R2 they want to go this way so to speak the current wants to flow in another direction rather than go into a resistor that has a very large value so anytime you increase R2 you're going to have a buildup of potential at this point because it's going to be very hard for for the electrons to pass through R2. So, at this point...
point you're going to have a higher potential. So by increasing R2 you increase the voltage across it and it makes sense because V equals IR. So if you can increase the resistance the voltage across that resistor is going to go up assuming the current doesn't change much.
Now let's talk about voltmeters and ammeters. An ammeter is used to measure the current that flows in a circuit. And a voltmeter is used to measure the volts.
Now the way you want to connect an ammeter, you want to connect the ammeter in series. with the circuit. You want the current to flow through the ammeter.
Therefore, since you want to measure the current, you want the ammeter to have as little as resistance as possible. So for example, let's say if this resistor has a value of 100 ohms and the voltage is 100 volts. Without the ammeter, the current that would flow through the circuit would be 1 amp.
V equals IR, 100 divided by 100 is 1. Now let's say the ammeter has a resistance, an internal resistance of 10 ohms How would that affect the accuracy of the current measured? So if the ammeter has a resistance of 10, the total resistance is not 110. So the current is going to be the voltage of 100 divided by the total resistance of 110. And so we won't get an accurate measurement. 100 divided by 110 is equal to about 0.91. one.
So the current measured will be less than the current measured. than the actual current. Now, let's say if the ammeter has an internal resistance of 1 ohm instead of 10 ohms. What current will it measure in this case?
So V equals IR, the current is the voltage divided by the total resistance, in this case, which is 101. So notice that we're going to have a more accurate measurement. If the resistance is 1 ohm, then the current... measured is 0.99.
Notice that it's closer to the true value of 1 amp. Now what if we decrease the internal resistance to, let's say, 0.1 ohms? Notice that it's going to be even more accurate. 100 divided by the total resistance of 100.1. This is about 0.999, which you can round that to 1. So as you can see, whenever you want to design an ammeter, you want the internal resistance of the ammeter to be extremely small, so that you can get a very accurate measure of the current in a circuit.
The smaller the internal resistance of the ammeter, the better. Now what about a voltmeter? How should we use it in a circuit?
So we said that for an ammeter, we want to connect it in series with the circuit, since you want to measure the current. And therefore, ammeters require a very low internal voltage. resistance now a voltmeter you want to use it to measure the voltage across an element so you don't want to connect it in series with the circuit you want to connect it across an element So let's say if this is 5 ohms and this is 10 ohms.
Actually, let's say it's 15 ohms. And the voltage across the circuit is 60. What is the voltage across the circuit? voltage across the 5 ohm resistor. So first let's find the current through it.
The total resistance is 15 plus 5 which is 20 and V equals IR, 60 volts divided by 20 ohms will give us a current of 3 amps. Now if you use V equals IR again, the voltage across the 5 ohm resistor is the 3 amps that flows through it times 5. So it's going to be 15 volts across the 5 ohm resistor. So let's say this is 0, this is going to be 15, and at this point it's going to be 60 volts.
Now the voltmeter typically has a black probe and a red probe. Now what voltage will it read if you connect, let's say, the black probe to the bottom of the 5 ohm resistor and the red probe at this point? Will it read positive 15 or negative 15?
If the red probe is at the higher voltage, it's going to read positive 15. Now, if the red probe... is at the lower voltage, it's going to read negative 15. So you're still going to get a potential difference of 15, but the sign tells you the direction. So if the red probe is at the higher voltage, you're going to get a positive value.
If the red probe is at a lower voltage, the meter is going to read a negative value. Now, as we said before, the ammeter requires a low internal resistance to accurately measure the current that flows in a circuit. Now, the voltmeter requires a very high resistance to accurately measure the voltage across an element.
Now, let's understand why. Let's use the... concept of the internal resistance to figure this out so let's say if we have a battery and we have the internal resistance of the battery and then we have this external resistance so this This represents the battery.
The resistor that's inside the red box is the internal resistance of the battery. The resistor on the outside is basically the internal resistance of the meter. So the blue circle represents the voltmeter Now let's say the internal resistance of the battery Let's say it has a value of 1 ohm. And the EMF, the voltage produced by the battery, let's say it's 12 volts. Now let's say that the meter has a resistance of 10 ohms how accurate will the meter be in reading the voltage So the meter is going to read the voltage between these two points.
So notice that we have a series circuit. Because the current that flows through the big R is equal to the current that flows through little r, those two resistors are in series, since the current is the same. Little r is the eternal resistance of the battery. So, to calculate the voltage, we need to find the current in the circuit. The total resistance is 1 plus 10, which is 11. So the current that...
flows through the circuit is the 12 volts the EMF which is also known as electromotive force or electron moving force you can also describe voltage that way it's the EMF divided by the total resistance so 12 divided by 11 is about 1.09 that's the current that flows through the circuit now the voltage across the 10 ohm resistor is going to be the current that flows through it which is 1.09 times the resistance of 10 so the voltage that the battery reads is 10.9 volts which is not very accurate But notice what happens if we increase the resistance of the meter. Let's say if we increase it to a thousand. So therefore, the total resistance of the circuit will now be 1001. So the current that flows through the circuit is the EMF of the battery, 12 volts, divided by the total resistance of 1001. So 12 divided by 1001 is going to be about 0.011988.
So now if we find the current, I mean if we find the voltage that is across the meter, it's going to be V equals IR. It's the current that flows through the meter, which is 0.011988 times the resistance of 1000. The voltage that the meter reads is going to be much higher. It's 11.988, which is approximately 11.99, and that's very close to 12. So by increasing the internal resistance of the voltmeter, meter we draw less current on the battery and if you draw away less current on the battery the voltage across the resistors going to be higher if you draw more current away from the battery you're putting a greater load on the battery, so its voltage is going to drop a lot, and you won't measure the actual voltage of the battery.
So notice that whenever the internal resistance of the meter is very high, the reading is more accurate. So that's why voltmeters have a very high internal resistance, so it can measure voltage properly. And the second reason is by increasing the internal resistance of the multimeter, or the voltmeter it draws less current, which puts a smaller load on the battery, so the voltage doesn't drop as much. Whenever the current in the circuit is high, the voltage drop is high, and so it's going to be difficult to measure the actual voltage of the battery.
Now let's work on some more circuit problems. Let's say if you have a circuit that looks like this. Let's say this is a 10 ohm resistor This is a 2 ohm resistor.
This is 3 and this is 5 and the voltage is 30 volts So what is the total current that flows through the circuit? And what is the current that flows through each resistor? So we need to simplify the circuit.
Notice that the current that flows through the 2 ohm resistor is the same. same current that flows to the three ohm resistor which is the same current that flows to the five ohm resistor because the current through those three resistors are the same those three resistors are in series with respect to each other which means that we can add them to get the equivalent resistance so we can simplify our circuit like this Whenever we have resistors in series, you can add them. 2 plus 3 plus 5 is 10. So we have two resistors that have a value of 10 ohms. Now notice that the current...
current splits off into multiple directions so we now have a parallel circuit and the voltage across the 210 ohm resistors is the same it's 30 they're connected across the same battery so we know that those two resistors are in parallel with respect to each other since they have the same voltage across each other now whenever you have two resistors in parallel and if they have the same value the equivalent resistance or the equivalent resistor is going to be half of what they were so it's going to be five and you can use the equation req is equal to 1 over r1 plus 1 over r2 raised to the minus 1. 1 over 10 plus 1 over 10 is 2 over 10. 2 over 10 raised to the negative 1, you flip the fraction, is 10 over 2. 10 divided by 2 is 5. So if you have two resistors to each other if they share the same value the equivalent resistance is half of one of those two resistors so now we can find the total current that flows to the circuit 30 divided by 5 is 6, so 6 amps of current leaves the battery. Now let's look at our second circuit. So we can see that 6 amps of current leaves the battery, and these two resistors, they have the same value.
That means that the current that flows through them is the same. So 3 amps is going to flow through the first 10 ohm resistor, and 3 amps will flow through the other one. So in our original circuit, we know that 3 amps will flow through the will flow through the first 10 ohm resistor and the other three amps will flow through the other three resistors which has an equivalent resistance of 10 ohms so now we have the current that flows through each resistor Now we can calculate the voltage of each resistor. By the way, we can redraw the circuit to make it look better and easier to work with. So here's the 10 ohm resistor and here's the other three resistors.
So you can view the circuit this way as well. So this is 30 volts. We have 6 amps of current leaving the battery. Here is the 10 ohm resistor. This is 2, 3, and 5. So we have 3 amps flowing through the 10 ohm resistor.
resistor and three amps flowing through the other three resistors so you can see that the voltage across the 10 ohm resistor is 3 times 10 which is 30 and You can also see that the voltage across the 2 ohm resistor is 2 times 3, which is 6 volts. Across the 3 ohm resistor, it's 3 times 3, which is 9 volts. And across the 5 ohm resistor, it's 3 times 5, which is 15 volts.
6 plus 9 plus 15 adds up to 30. So everything is balanced in the circuit. Now let's try another example. So this time we're going to have more resistors.
So let's say... This is going to be a 4 ohm resistor. This is 5. This is 6. This is going to be 15. And this is going to be 2.5 and 10. and let's say that the voltage is 40. Go ahead and calculate the current that flows through the circuit.
So first, let's combine... these three resistors into a single resistor because the current that flows through them is the same so they're connected in series 4 plus 5 plus 6 is 15 so therefore we have a circuit that is equivalent to this circuit So this is 15, and this is 15. So we have two resistors in parallel, and they have the same value. So we can combine those two resistors as a single resistor with half the value.
So 1 over 15 plus 1 over 15 is 2 over 15. And raised to the minus 1, that's 15 over 2, which is 7.5. So therefore, this resistor has an equivalent value of 7.5. And it's in series with a 2.5 resistor and a 10 resistor. So we can add these three.
2.5 plus 7. 7.5 is 10, 10 plus 10 is 20, so the equivalent resistance of the circuit is 20 ohms, and the voltage across the battery is 40. So V equals IR, 40 divided by 30. 20 is 2, so 2 amps of current leaves the battery. That means 2 amps of current flows through the 10 ohm resistor. So what is the voltage across the 10 ohm resistor?
Now let's say that the voltage here is 0. It might be good to define the absolute voltages at every point. So we know the voltage here is 40. So if 2 amps flow through the amplifier, through the 10 ohm resistor using V equals IR 2 times 10 is 20 so we have a voltage drop of 20 so 40 minus 20 is 20 so the voltage at this point is 20 volts now that we have that what is the current that flows through the 15 ohm resistor and through the 4 5 and the 6 ohm resistor keep in mind these three resistors had a value of 0. value of 15 which is equivalent to 15 as well therefore because this section and this section has the same resistance the current that flows through those two sections are the same so one amp of current flows through the 15 ohm resistor and one amp of current flows through the other three resistors 15 One times one is fifteen. 4 plus 5 plus 6, which is 15, times 1 is 15. So the voltage drop is the same.
So the voltage at this point is going to be 15 less than 20. So it's 5 volts. And so we can find the current across the 2.5 ohm resistor. We have a voltage difference of 5. 5 divided by 2.5 is 2. So 2 amps of current is flowing back to the battery. If two amps leave the...
the battery, 2 amps must flow back to the battery. Now let's find the other voltages. So if we have 1 amp flowing through a 4 ohm resistor, 1 times 4 is 4, so the voltage drop is 4, which means the voltage here is 16. 16 minus 5 is 11, and 11 minus 6 gives us 5. So we can see that the voltages, they all add up, so to speak.
So now let's use Kirchhoff's voltage law to see that the sum in any loop is always 0. So let's focus on this loop. So let's start at this point. Going up through the battery, the voltage is going to increase by 40 as we go from 0 to 40. So the battery is going to pump up the voltage from 0 to 40. Next, as we... As we go through the resistor, we experience a voltage drop from 40 to 20, so it drops by 20. And as we go through the 15 ohm resistor, the voltage drops by 20. voltage decreases from 20 to 5 so it decreases by 15 that's the voltage drop across the 15 ohm resistor and going back to where we started it goes down from 5 to 0 so we have a voltage drop of 5 40 minus 20 minus 15 minus 5 is 0 so now let's look at the next section Let's start, let's look at this part. So let's start at this position, and let's travel in this direction.
So as we go from 5 to 20, the voltage increases by 15. next as we travel from 20 to 16 the voltage drops down by four so we have a voltage drop and as we travel from 16 to 11 it drops by five and as you go from 11 Back to five it drops by six Fifteen minus four is eleven eleven minus five is six six minus six is zero So that works now there's another cycle or another section that we can look at and that's for the This outer region So let's start at this point again, and let's go around the circle. So as we go up through the battery, it's going to increase the voltage by 40. It went from 0 to 40. Now, starting from 40, we're going to go to 20, so that's a voltage drop of 20. And then from 20, we're going to travel to 16, so that's a voltage drop of 4. And from 16, we're going to go to 11, so that's a voltage drop of 5. and from 11 we're going to go to 5 that's a voltage drop of six and from five we're gonna go back to zero which is a voltage drop of five so 40 minus 20 is 20 20 minus 4 is 16 16 minus 5 is 11 11 11 minus 6 is 5, 5 minus 5 is 0. So as you can see, if you solve the circuit correctly, the voltage as you go around any loop will always add to 0. Now consider this circuit. Let's say if we have two resistors in series, and let's say this is a 3 ohm resistor, and this is a 7 ohm resistor, and the voltage is 40. The total resistance is 10, and so the current that flows through the circuit is 40 over 10, which is 4 amps.
Now let's say if we add another battery. like this and it's the same circuit so if we connect the negative side of one battery to the positive side of the other battery and if we use the same resistance, let's say this is 3 ohms and this is 7 ohms, and the first battery is 40 and the second one has a voltage of 10, will the current that flow through the circuit, will it be greater than 4 or less than 4? When you connect the positive terminal of one battery to the negative terminal of the other battery, the voltages are additive.
Both batteries are delivering power to the circuit, so both batteries are in discharge mode. They're delivering energy to the circuit, which means that they're losing energy. Over time, the batteries will run out of juice.
So, if we draw the direction of the current, it's going to be in this direction. So whenever the current leaves the positive terminal of the battery, the battery will battery is delivering energy so the total voltage of the circuit is 50 volts and now let's create a map voltages at every point. So over here the voltage is 0, then this side is higher by 10 because this is the positive terminal. So at this point the voltage is 10. Here the difference is 40 and the positive side means that it's 40 volts higher than the negative side, so this side is 50. So the voltage across the two resistors is 50 volts 50 minus 0 so we have 50 volts volts across an equivalent resistance of 10 so the current that flows through is 50 divided by 10 which is 5 amps so therefore the voltage across the 3 ohm resistor is 5 times 3 which is 15 volts and across the 7 ohm resistor is 5 times seven that's 35 volts so the voltage here is 35 so if you go from 50 to 35 that's a voltage drop of 15 and if you go from 35 to 0 that's a voltage drop of 35 Now let's say if we connect the batteries in a different way. Let's say if we connect the negative terminal of one battery to the negative terminal of the other battery.
What's going to happen now? Let's say this is still 3 ohms, and this is 7 ohms, and this is 40 volts, and this is 10 volts. So what I would do is identify which side is positive and which side is negative.
which side is negative now it helps if you identify the absolute voltages of every point across the circuit so we're going to say this is 0 volts now this is 0 volts what is the voltage here notice that this side is positive which means that that side is 10 more than this side so this side is less because it's negative that means that it's at negative 10 volts now the 40 volts means that this side is 40 volts higher than negative 10 so negative 10 plus 40 is 30 So the voltage here is 30 volts. So as you can see, these two batteries they're not working together they're going against each other the battery with the higher voltage is charging the battery with the low voltage so the 40 volt battery is supplying power to the circuit. It's losing energy as it delivers power.
The 10 volt battery is absorbing some of the energy that the 40 volt battery is supplying. So the 40 volt battery is in discharge mode. it's going to die out over time. The 10 volt battery is in charging mode. So this circuit will continue and the voltage of the 40 volt battery will drop as it loses energy and the voltage of the 10 volt battery will go up as it charges.
Eventually the two voltages will meet as the 40 volt battery decreases and the 10 volt battery increases in power. There's going to be a point where the voltage will go up as it charges. So the voltage of the 10 volt battery will go up as it charges. So the voltage of the 10 volt battery will go up as it charges. So the voltage of the voltage is the same and at that point no current will flow these two batteries will be at a standstill but at this moment we have a potential difference of 30 volts across the equivalent 10 ohm resistor so therefore the current that flows through this circuit is 30 divided by 10 which is 3 amps so the voltage across the three arm resistor is 3 times 3, so that's 9 volts, and the voltage across the 7-ohm resistor is 7 times 3, which is 21. So therefore, the voltage here is 21 volts.
So if we travel around the circuit, going this way, the voltage decreased from 0 to negative 10. Going further, it increases from negative 10 to 30, so it goes up by 40. And then going this way, it decreases from 30 to 9, so we have... Not 30 to 9, I'll take that back. 30 to 21. So we have a voltage drop of 9, so that's negative 9. And then from 21, it decreases to 0, so we have a voltage drop of negative 21. Negative 10 plus 5. 40 is negative 30 now I take that back negative 10 plus 40 is positive 30 positive 30 minus 9 is 21 and 21 minus 21 is 0 so in all cases the voltage sum must be 0 anytime you go in a complete circle around the circuit now the last thing that I do want to mention is the difference between DC current and AC current.
DC current is direct current. This is current that flows in one direction. Batteries provide such current.
So in a DC circuit, you have current flowing in a single direction. But in an AC circuit, which stands for alternating current, AC current Constantly changes direction. At one point the current's going this way, then it reverses and so it's going that way.
And because it constantly reverses, the electrons in the circuit are moving back and forth. And so there's no net movement in the AC circuit. If you plot the voltage versus time for a DC circuit, it's going to look like this. It's simply a straight line. The voltage is constant.
But for an AC circuit, the voltage varies based on a sine wave. Now the frequency is the number of times the current changes direction for an AC circuit. So if the frequency is 60 Hz, that means it changes direction 60 times in a single second.
So I just want to give you just an intro into AC current and how it relates to DC current. But I'm going to stop here for this video. Hopefully you found it to be educational, so thanks for watching, and have a great day.