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Has the current through battery 1 changed?

  • Thread starter cy19861126
  • Start date
  • #1
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Homework Statement


In this problem, box A and box B contain unknown combinations of light buls. Bulb 1 is identical to bulb 2. The batteries are ideal.
1. a) In the circuit A (see first attached pic) the voltage across bulb 1 and the voltage across box A are equal. What, if anything can you say about the resistance of box A compared to the resistance of bulb 1. Explain
b) Write an expression for the voltage across the battery, V(Bat), in terms of the voltage readings across box A and across the bulb V(Box A), V (Bulb 1)
2. a) In the circuit B (see second attached pic) the voltage across bulb 2 and the voltage across box B were find to be equal. What, if anything, can you say about the resistance of Box B compared to the resistance of bulb 2?
b) Write an expression for the voltage across the battery V(Bat) in terms of the voltage readings across box B and across the bulb V(Box B) V (Bulb 2)?
3. Box A and box B are now interchanged. It is observed that bulb 2 is now brighter than it was when box B was in that circuit
a). Is the resistance of box A greater than, less than, or equal tot he resistance of box B?
b) Has the current through battery 1 changed? If so, how?
c) Has the current through battery 2 changed? If so how?



Homework Equations


V = IR
Series: Req = R1 + R2
Parallel: Req = 1/R1 + 1/R2


The Attempt at a Solution


1. a)Nothing. Without current given, we pretty much cannot tell which resistence is higher.
b) V(Bat) = (V(BoxA)+V(bulb1))/2
2. a) The resistence are the same
b) V(Bat) = B(Box B) + V(Bulb 2)
3. a) Less than. Brighter bulb means more current and less resistence
b) Yes, more current
c) Yes, less current

PS: It seems like I have no problem in determining current and resistence from the given information, but a huge problem in determining voltage. Please provide some clue to this problem. Many thanks~~
 

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Answers and Replies

  • #2
ranger
Gold Member
1,676
1
Well the voltage in the first circuit would be the same as the source voltage because all components are in parallel. So thats no problem.

As for the second circuit series circuit, if you have equal resistances, then the voltage is shared equally by the bulbs, which you already stated. However upon changing boxes, you find the brightness of the bulb increases. This means that the current increases in the circuit. So given the equation V = IR, does that help in anyway to give you an idea about the change in voltage? Remember that the current increased due to increased brightness, therefore for the bulb, we have more current flowing through it which gives it a.....?
 
  • #3
69
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From your statement, it seems that voltage and current are directly related. Increasing the current increases the voltage by using the equation, V = IR. However, there is also this other equation P = IV. This equation shows you that current and voltage are inversely related. How can this be so?
 
  • #4
ranger
Gold Member
1,676
1
Indeed, voltage is proportional to current as given by V = IR. When you consider the case that you've asked, it is true that V = P/I. But power can be either IV, I2R, or V2/R. So if we can say
V=P/I = (IV)/I = V
V = (I2R)/I = IR = V
V = (V2/R)/I = V2/(IR) = V2/V = V

You will find the same is true for current. What I'm getting at here is that we cant have power without a combination or V,I, or R. And no matter what the combination, we get V = V which is the same as V = IR, so we come back to the direct relation of V and I.
 

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