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 kum and T is the time in seconds the unit for current is the amp so 1 amp is equal to 1 km per second an electric charge is associated with the quantity of charged particles an electron has a charge that's equal to 1.6 * 10^9 kum 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 I 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 will 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 sevenlane highway it's going to be easy for cars to flow through it as opposed to a onelane highway a onelane 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 I R you can get the second form which is I 2 * R and if you replace I with V / R you can get the third form which is V ^2 R so power is equal to voltage time current or I^2 * R or V ^2 R power is measured in watt power is the rate at which energy can be transferred 1 watt is equal to 1 jewel 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 this 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 / T so to solve for the electric charge Q it's I * 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 got to multiply by 60 notice that the unit minutes cancel 12 * 60 is uh 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 2,7 36 kms now what about part B how many electrons would this represent so if you have the charge you can easily convert it to number of electrons let's start with uh this number now it turns out that one electron has a charge of 1.6 * 109 we don't have to worry about the negative sign so if we divide these two numbers 2736 / 1.6 6 * 109 this will give you the number of electrons and so that's going to be about 1.71 * 10 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 9V 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 9volt 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 volt / 250 ohms is equal to 036 amps now if you want to convert amps into milliamps multiply by th00and 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^2 * R the current that flows through the resistor is 036 amp and we need to square it and resistance is 250 ohms 036^ squar is 01296 and if we multiply that by 250 this is going to give us a 324 watts which is equivalent to 324 m part C how much power is delivered by the battery well let's use this equation p is equal to V * 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 036 amps 9 *.36 is equal to the same thing 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 that's the 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 12volt 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 15 amps so now we could find the electrical resistance using Ohm's law v is equal to IR so the voltage is 12 the current is 15 and let's solve for R we can do that by dividing both sides by 15 so the electrical resistance is equal to the voltage / the current 12 / 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 15 so 12 *.15 that's uh 1.8 watt now we can also use I^ 2 * R so we can take the current which is5 squared and then multiply by the resistance which is 80 15^2 * 80 will give us the same answer 1.8 W 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 W 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 watt and not 80 W so let's go ahead and begin with that so first we need to convert watts into kilowatt 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 kilowatt let's divide by a th00and there's 1,00 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 is about 24 hours per day so notice that the unit months cancel and the unit days cancel as well leaving us with kilowatt time 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 /,000 * 30 * 24 *.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 got to divide it by 1000 which is 4 amp so what is the voltage well electrical power is equal to voltage time current so P is 50 we're looking for V and I is4 so we need to divide both sides by 04 so 50 /4 is 125 so that's the voltage it's 125 vol 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 is4 and let's find R so let's divide both sides by 04 125 /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= R they're very useful in solving common problems number five 12.5 of charge flows through a 5 koh resistor in 8 minutes what is the electric current that flows through the resistor so we have the charge Q it's 12.5 kms and we have the electrical resistance which is uh 5 kiloohms 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 it's 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 * 8 is 48 so 60 * 8 is 480 you just got to add the zero so now we can find the electric current the charge that flows through any given point is 12.5 kum now let's divide that by 480 seconds keep in mind 1 amp is 1 km per second 12.5 / 480 is 0.26 kum/ second or simply 026 amps which is equivalent to 26 milliamps now let's calculate how much power is consumed by the resistor so let's uh 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 026 amps and let's square that number and the resistance was 5 kiloohms a kiloohm is 1,00 ohms so 5 kiloohms is 5,000 ohms 026^ 2 is very small like 6.76 * 10 - 4 and if we multiply that by 5,000 this is going to equal 3.38 watt 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 * R it's 026 amps times the resistance of 5,000 and that's about 130 volts so that's the voltage across the resistor and this is the electrical power which is what we want