In this video, we're going to go over a few basic equations and work on some practice problems involving electric current and Ohm's law. So let's say if we have a battery. The long side of the battery is the positive terminal.
And here's a resistor. Conventional current states that current flows from the positive terminal to the negative terminal. Current flows from high voltage to low voltage. That's conventional current, the same way as water flows from a high position to a low position.
Electron flow is the opposite. In reality, we know that electrons, they emanate from the negative terminal and flow towards the positive terminal. So just keep that in mind. But now let's talk about current, conventional current, which is the flow of positive charge.
Current is defined as, it's basically the rate of charge flow. It's charge divided by time, or delta Q over delta T. Q is the electric charge measured in coulombs, and T is the time in seconds.
The unit 4 current is the amp. So 1 amp is equal to 1 coulomb per second. An electric charge is associated with the quantity of charged particles. An electron has a charge that's equal to 1.6 times 10 to the negative 19 coulombs, and it's negative. Now, there are some other equations that we need to talk about, and one of them is Ohm's Law, which describes the relationship between voltage, current, and resistance.
V is equal to IR. Voltage is the product of the current and resistance. The resistance is measured in ohms.
That's the unit of resistance. Now, keeping the resistance constant, if you were to increase the current, what do you think the effect would be on the voltage? Or rather, if you increase the voltage, what is the effect on the current?
Increasing the voltage will increase the current. And what about increasing the resistance? What effect will that have on the current? If you increase the resistance, the current will decrease.
But if you were to increase the voltage, the current will increase. The voltage and the current are directly related. The resistance and the current are inversely related. The more resistance you have in a circuit, It's harder for current to flow. It's just, it's not going to flow as well.
Think of a highway. If you have a seven lane highway, it's going to be easy for cars to flow through it. As opposed to a one lane highway. A one lane highway has more resistance, so less cars can flow through it. The cars being the flow of electric current.
But if you decrease the resistance, if you add more lanes to the road, more cars can flow, so there's more current. The next equation you need to be familiar with is electric power. Electric power is the product of the voltage and the current. Now, there's three forms to this equation. So, if you replace V with IR, you can get the second form, which is I squared times R.
And if you replace I with V over R, you can get the third form, which is V squared over R. So, power... is equal to voltage times current, or I squared times R, or V squared over R. Power is measured in watts.
Power is the rate at which energy can be transferred. One watt is equal to one joule per second. So these are some things to know. Now let's work on some problems.
A current of 3.8 amps flows in the wire for 12 minutes. How much charge passes through any point in the circuit during this time? So we have the current, it's 3.8 amps, and we have the time, which is 12 minutes. How can we calculate the electric charge?
Well, we know that I is Q divided by T. So to solve for the electric charge Q, it's I times T. So we have to multiply, but we need to be careful with the units though.
T is the time in seconds. So let's convert 12 minutes into seconds. Each minute equates to 60 seconds.
So we've got to multiply by 60. Notice that the unit minutes cancel. 12 times 60 is 720. So T is 720 seconds. Now let's calculate Q. So it's equal to I, the current, which is 3.8 amps, multiplied by 720 seconds.
So the electric charge is 2,736 coulombs. Now what about part B? How many electrons would this represent? So if you have the charge, you can easily convert it to the number of electrons. Let's start with this number.
Now, it turns out that one electron has a charge of 1.6 times 10 to the negative 19 coulombs. We don't have to worry about the negative sign. So, if we divide these two numbers, 2736 divided by... by 1.6 times 10 to negative 19 this will give you the number of electrons and so that's going to be about 1.71 times 10 to the 22 electrons so keep in mind the amount of charge is proportional to the number of electrons so that's it for this problem Number two, a 9-volt battery is connected across a 250-ohm resistor.
How much current passes through the resistor? Well, we can begin by drawing a circuit. Here's the battery, and here is the resistor. So we have a 9-volt battery and a 250-ohm resistor.
What equation do we need to calculate the electric current? The equation that we can use is Ohm's Law. V is equal to IR.
The voltage is 9, and the resistance is 250. So, solving for I, let's divide both sides by 250. So the current is equal to the voltage divided by the resistance. So 9 volts divided by 250 ohms is equal to 0.036 amps. Now, if you want to convert amps into milliamps, multiply by 1,000. Or move the decimal three units to the right. So this is equivalent to 36 milliamps.
Part B. How much power is dissipated by the resistor? So what equation can we use here? Well let's use this equation power is equal to i squared times r.
The current that flows through the resistor is 0.036 amps and we need to square it. And the resistance is 250 ohms. 0.036 squared is 0.001296 and if we multiply that by 250, this is going to give us 0.324 watts.
which is equivalent to 324 milliwatts. Part C, how much power is delivered by the battery? Well, let's use this equation, P is equal to V times I.
The voltage of the battery is 9 volts, and the current that the battery delivers is the same as the current that flows through the resistor, which is 0.036 amps. 9 times 0.36 Thank you. is equal to the same thing, 0.324 watts. And it makes sense. Everything has to be balanced.
The amount of power delivered by the battery should be equal to the amount of power dissipated or absorbed by the resistor. Because there's only two elements in a circuit. The battery delivers energy, the resistor absorbs it. So if they're the only two circuit elements, the amount of power transferred has to be equal.
Number three, a 12-volt battery is connected to a light bulb and draws 150 milliamps of current. What is the electrical resistance of the light bulb? Let's draw a circuit.
So let's say this is the light bulb, and we have a battery connected to it. And that's the positive terminal, here's the negative terminal. And electric current flows from the positive side to the negative side. But electrons will flow in the opposite direction.
Now let's make a list of what we know. So the voltage is 12. The current is 150 milliamps. But we need that in amps.
So we can divide that by 1000 or move the decimal 3 units to the left. So that's equivalent to 0.15 amps. So now we can find the electrical resistance using Ohm's law.
V is equal to IR. So the voltage is 12, the current is 0.15, and let's solve for R. We can do that by dividing both sides by 0.15. So the electrical resistance is equal to the voltage divided by the current.
12 divided by 0.15 is 80. So the internal resistance of the light bulb is 80 ohms. Now how much power does it consume? Well, we can use P is equal to VI. The voltage across the light bulb is equal to the voltage of the battery.
That's 12. And the current delivered by the battery is equal to the current absorbed by the resistor. So that's 0.15. So 12 times 0.15, that's 1.8 watts.
Now we can also use I squared times R. So we can take the current, which is 0.15 squared, and then multiply by the resistance, which is 80. 0.15 squared times 80 will give us the same answer, 1.8 watts. So you can use both techniques to calculate the electrical power.
Now what about part C? How much will it cost to operate this bulb for a month if the cost of electricity is 11 cents per kilowatt hour? Well, we know the power that it uses is 80 watts. Let's find out how much energy it uses in a month, then we can find out the cost. Now I do have to make a small correction.
The power is 1.8 watts and not 80 watts. Now let's go ahead and begin with that. So first we need to convert watts into kilowatts. We need to find the energy in kilowatt hours.
Energy is basically power multiplied by time. Power is energy over time. Electric power is the rate at which energy is transferred. Now to convert watts to kilowatts, let's divide by a thousand.
Here's a thousand watts per kilowatt. Now we need to multiply by the number of hours. The total time that the light bulb is going to be operating is for one month. And there's 30 days in a month on average. And there's about 24 hours per day.
So notice that the unit months cancel. and the unit days cancel as well leaving us with kilowatt times hours so this will give us the amount of energy being consumed in one month to find the cost let's multiply by 11 cents per kilowatt hour and so now the unit kilowatts will cancel and the unit hours will cancel as well So now all we need to do is just the math. So it's 1.8 divided by 1,000 times 30 times 24 times 0.11. So it's only going to cost 14 cents to operate the light bulb for a month.
Number four, a motor uses 50 watts of power and draws a current of 400 milliamps. What is the voltage across the motor? So we have the power, which is 50 watts.
It's always good to make a list of what you have. And the current... is 400 milliamps. We want that in amps.
So we've got to divide it by 1000, which is 0.4 amps. So what is the voltage? Well, electrical power is equal to voltage times current.
So P is 50, we're looking for V, and I is 0.4. So we need to divide both sides by 0.4. So 50 divided by 0.4 and Is 125. So that's the voltage.
It's 125 volts. Now what about part B? What is the internal resistance of the motor?
Well, let's use Ohm's Law. V is equal to IR. So V is 125. I is 0.4.
Now let's find R. So let's divide both sides by 0.4. 125 divided by 0.4. is equal to 312.5 ohms.
So as you can see, these two equations are very important. P is equal to VI and V equals IR. They're very useful in solving common problems. Number five.
12.5 coulombs of charge flows through a 5 kiloohm resistor in 8 minutes. What is the electric current that flows through the resistor? So we have the charge queue, it's 12.5 coulombs, and we have the electrical resistance, which is 5 kilo ohms, and we have the time, 8 minutes.
How can we use this information to calculate the electric current? Well, the electric current is the ratio, or really it's the change in the electric charge, divided by the change in time. So, it's the rate of charge flow. how much charge flows per second which means that we need to convert 8 minutes into seconds so we got to multiply it by 60 seconds 6 times 8 is 48 so 60 times 8 is 480 you just gotta add the zero.
So now we can find the electric current. The charge that flows through any given point is 12.5 coulombs. Now let's divide that by 480 seconds. Keep in mind, 1 amp is 1 coulomb per second.
12.5 divided by 480 is 0.026 coulombs per second, or simply 0.026 amps. which is equivalent to 26 milliamps. Now let's calculate how much power is consumed by the resistor. So let's make some space before we do that. What equation would you use?
Now we don't have the voltage so let's use an equation that contains only current and resistance. The current is 0.026 amps. Let's square that number and the resistance was 5 kilo ohms.
A kilo ohm is 1000 ohms so 5 kilo ohms is 5000 ohms. 0.026 squared is very small. It's like 6.76 times 10 to the minus 4. And if we multiply that by 5000, this is going to equal 3.38 watts. So that's how much power is consumed by this resistor.
If we want to find the voltage, we can. It's simply equal to I times R. It's 0.026 amps times the resistance of 5000. That's about a hundred and thirty volts, so that's the voltage across the resistor, and this is the electrical power Which is what we want?