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 per second and 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 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 high wing 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 resistance if you add more lanes to the road more cars could 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 this 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 high 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 2736 columns 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 this number now it turns out that one electron has a charge of 1.6 times 10 to the negative 19 coulombs i'm gonna have to worry about the negative sign so if we divide these two numbers 27 36 divided by 1.6 times 10 to the 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 nine 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 a thousand 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 .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 is equal to the same thing point 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 the 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'll need that in amps so we can divide that by a thousand or move the decimal three units to left so that's equivalent to point 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 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 uh 1.8 watts now we can also use i squared times r so we can take the current which is point fifteen square it and then multiply by the resistance which is eighty point fifteen squared times eighty 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 now 80 watts so 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 there'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 of 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 a thousand 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 got to divide it by a thousand 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 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 i r so v is 125 i is 0.4 and 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 twelve point five columns of charge flows through a five kilo ohm resistor in eight minutes what is the electric current that flows to the resistor so we have the charge q it's 12.5 coulombs and we have the electrical resistance which is uh 5 kilo ohms and we have the time eight 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 eight 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 and let's divide that by 480 seconds keep in mind one amp is one clone per second 12.5 divided by 480 is .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 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 0.026 amps and let's square that number and the resistance was 5 kilo ohms a kilo ohm is a thousand ohms so five kilo ohms is five thousand ohms point zero two six squared is very small it's like six point seven six times ten to the minus four 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 .026 amps times the resistance of 5000. that's about 130 volts so that's the voltage across the resistor and this is the electrical power which is what we want