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 (a) the current through each bulb.

Video transcript

Hey everyone. So today we're dealing with the problem about circuits. So we're being told that we have to heating coils that have resistance is of 500 homes. And 700 homes were also being told that they're powered by a 110 volt power source with this information, we're being asked to determine the current and the heating coils when connected in series. Now all of our answer choices give us the current over the 500 ohm resistor and the 700 ohm resistor, which I'll call R one and R two respectively are one is homes and our two Is 700 homes. Now while all the answer choices give us currents separately for both. One thing that we should take note of is that the system is in series and in series current is the same through all resistors connected in series, same through so therefore the current will be the same through R one and R two. Therefore therefore we can also write that The current through 500 will be equal to, I threw with that in mind. We can also take a look at Homes law because we're going to need it for this problem, which gives us the relation between voltage current and resistance with this in mind. Let's also go ahead and draw our circuit just to sort of conceptualize this a little better. So we have our voltage source which is 100, Connect to our one R one and our two, they're connected in series. So they're in line with each other. So since the current through R one and R two are the same, then this must mean that the current through the system, the entire system will be our answer as well. So to do this we need to find the equivalent or total resistance in the system, which since our resistors are in series we can simply add them up Which gives us 500 plus With a total of 1200 homes. So this is our total resistance. We know our voltage is 110V. So rearranging OEMs law to solve for I the current in amperes we get that i is equal to V over R. Which will just be 110V over 1200V giving a final current of oops Of 0. and pierce. And as we mentioned earlier the current between the two resisters of the current that passes through each resistor will be the same Has the total here. So this means that the current in the heating coils when connected in series will be 0.0917 through both resistors or answer choice. A. I hope this helps. And I look forward to seeing you all in the next one