1. The problem statement, all variables and given/known data A circuit consists of three identical lamps connected to a battery. Battery has some internal resistance. Switch S, open at first, is closed. 1) What happens to brightness of lamps B and C? 2) What happens to current in battery? 3) What happens to potential across lamps A and C? 4) What happens to total power delivered to lamps by battery? See attached figure. 2. Relevant equations I = emf/R+r P = I*deltaV 3. The attempt at a solution This is my attempt at the solution; I am almost certain that my answers are correct but I would appreciate if someone could verify, and explain if possible, if I am incorrect. 1) B brightness remains the same. C brightness increases since the circuit is complete. 2) Current in battery decreases because of increased resistance. 3) Potential is unchanged for lamp A but increases for C because circuit is now closed. 4) Total power increases, since there is an increase in potential due to the closed circuit. I would appreciate any assistance, thank you!
You should consider what the effect will be of the internal resistance of the battery when the load changes. This may have some affect on your answers. You might also want to reconsider your reason for the increase in total power delivered; surely the battery's available potential is not increased? Can you think of what has increased or decreased with the closing of the switch?
P=P[R]+P[r]=(I^2R)+(I^2*r) I am having a hard time understanding how the variables in the system relate. I see know that the potential does not increase, though I believe that there will be equal potential on both sides of the circuit. There is an increase in current; per the above equation, does this imply that the reason for increased power is increased current?
There is increased current, yes. This is due to the addition of a new path for current being added when the switch is closed. Note that the "new" light bulb, C is effectively in parallel with the existing combination of A and B. How do you suppose this will effect the net load resistance?
well, because there is another element that adds resistance, that would lead to an increase in net resistance. increased load resistance decreases the internal resistance, r. because there is some fixed current I and constant (emf): I = emf/R+r an increase to the external load, R, would lead to a decrease in r. Is this analysis correct? Since the original question was: 2) What happens to current in battery? and there is additional current flowing through the system, there would be an increase in current flowing through the battery? i just realized that these statements are somewhat contradictory. can you also provide advice on how I should consider the system (i.e., each circuit independently, or together)?
Have you discussed in class how resistances combine in series and in parallel? You should consider, for the purposes of this problem, that both the emf of the battery and the internal resistance of the battery have fixed values. What changes is the voltage dropped across the internal resistance when the load current changes.
not yet, though I know resistance in parallel 1/R[eq] = 1/R[1]+1/R[2] ... etc. i dont understand how this applies to the system though, or 2) What happens to current in battery? 4) What happens to total power delivered to lamps by battery? (i am pretty certain the other answers are correct). i dont understand how to quantify the resistance in the system, does each bulb have a resistance R?
It's fair to suppose that each bulb will have some fixed resistance, and you wouldn't go wrong if you were to suppose that they all had the same resistance, say R. So.. Note that bulbs A and B are in series, so they represent a total resistance 2R. Bulb C, when the switch is closed, will place another resistance R in parallel with the 2R. Qualitatively, then, you can say that when the switch is closed the net load resistance that the battery sees will be lower than when the switch is open. More total current should flow through the lower resistance. Of course, with more total current flowing from the battery, there will be more current going through its internal resistance -- hence it should drop a bit more voltage. The "outside world", the circuit external to the battery, will see a slightly lower voltage due to this additional drop across the internal resistance. How do suppose this will alter the current through bulbs A and B?
Be a good idea if you read questions more carefully. You have answered (correctly) a part of the question based on the switch OPENING, but that wasn't the question.
so decreased voltage due to increased current flowing through the battery because of the decreased external resistance, as related by: deltaV=(emf)-I*r would lead to decreased current for light bulbs A and B per: I=deltaV/R[eq] but I can also argue that there would be increased current due to decreased external resistance R. I am still not getting it....
Yes, there is an increase in overall current. Consider that the decrease in load resistance is due to an additional path for the current to flow through (bulb C). The current through this new path more than makes up for the minor decrease in current via bulbs A & B.
I am going to analyze the problem by question because I am having a hard time putting this whole thing together. 1) What happens to brightness of lamps B and C? Brightness of lamp B decreases because of decreased potential, and thus, current: I=deltaV/R[eq] Brightness of lamp C increases because of closed circuit and increased current. 2) What happens to current in battery? The current through the battery increases because resistance in the overall system decreases. 3) What happens to potential across lamps A and C? Potential across lamps A and C decreases because of increased current through the battery: deltaV=(emf)-I*r 4) What happens to total power delivered to lamps by battery? Power increases since current through system and battery increases. P=P[R]+P[r]=(I^2*R)+(I^2*r) Is this correct? I am confused about #4, how do I relate the decreased external resistance to the increase in current through the system (from the closed circuit)?
I didn't verify your equations, but you have the word statements correct. I don't see why you have an issue with #4 since you have the right idea, unless you are just asking whether the equation you used is correct (sorry, I don't do equations much any more).
Okay. You might want to indicate that the decrease in potential is due to an increase in the voltage dropped by the battery's internal resistance. Okay. You might want to refer to the load resistance; It's the overall load resistance that has decreased. This one's a bit tricky because before the switch is closed bulb C has no potential across it. So technically it's potential increases when the switch is closed. You're right about lamp A, of course. Yes, total power delivered by the battery increases because of the lowered load resistance and resulting increase in current passing through bulbs (always presuming that the load resistance remains higher than the internal resistance of the battery).