Light Bulbs in Series and in Parallel. Two light bulbs have constant resistances of 400Ω and 800Ω. If the two light bulbs are connected in series across a 120-V line, find (c) the total power dissipated in both bulbs.

Video transcript

Hey everyone today, we're dealing with the problem about circuits. So here we're being asked to determine the power dissipated by two resistors or one. And our two which have uh resistance is of 115 and 145 homes respectively. Now the resistors are connected in series and powered by a 110 volt power source. So let's sort of draw this out to get an idea for it. Right? So here we have Our power source which is 110V. It's connected in series two and I'll draw this in blue one resistor, oops 21 resistor and it tells us that they are in series, which means there will be a second resistor which will be in line with it. It's not parallel. And then this will go back to the power source. So let's do that. We'll say this is our one Which is equal to homes And this is our two which is 145 homes. Now, if you recall power in the circuit can be defined with a simple equation. Power is equal to V into i where V is voltage, I is current and P is power in watts can also be written as I squared R. Where are as resistance in OEMs which can also be rewritten as b squared R. Or b squared divided by ar. Now all three of these are equivalent formula. So we can use any of them. We also need to remember homes law which states that voltage is equal to the product of current and resistance. Now taking a look at our circuit a little more in depth. We know that voltage across the resistor in series is less than the potential difference supplied by the power source and the power source voltage is applied across the coolant resistors. Therefore in a series circuit resistance or total resistance, let's write this in blue. Since we're dealing with blue for resistance, equal in resistance will simply be equal to the sum of all resistance is in a system provided that all the resistors are in series. So using the given values that we have so far, we know that our one is equal to 1 15 homes and R two is equal to 45 homes. Therefore the total resistance in the system, excuse me will be 260 Homes, 260 Homes. Now the potential difference, I cried. Oh sorry, applied across the equivalent resistors is 110V because again we're on series and that is the voltage supplied by the power source. So we can use that value as well. And now I'll do it in red for power. So power will be equal to the voltage squared divided by equal and resistance substituting in our values, we get that this is equal to 110 volts squared divided by 260 Homes, 260 Homes. And when we solve that, we get a final answer of 46. watts or answer choice B. Therefore The power dissipated by both resistors in the circuit is 46.5 watts. I hope this helps, and I look forward to seeing you all in the next one.