Two identical bulbs and a capacitor are connected to an ideal battery as shown.
ID: 1283404 • Letter: T
Question
Two identical bulbs and a capacitor are connected to an ideal battery as shown. The capacitor is initially uncharged.
a. Just after the switch is closed:
* What is the potential difference across bulb A, across bulb B, across bulb C and across the battery? Explain.
* Rank the currents iA, iB, iC and iBat, Explain your reasoning.
b. A long time after the switch is closed:
* Rank the currents iA, iB, iC and iBat, Explain your reasoning.
* What is the potential difference across bulb A, B, across the capacitor C and the battery? Explain.
c. Summarize your results by describing the behavior of bulb A and of bulb B from just after the switch is closed until a long time later.
Two identical bulbs and a capacitor are connected to an ideal battery as shown. The capacitor is initially uncharged. a. Just after the switch is closed: * What is the potential difference across bulb A, across bulb B, across bulb C and across the battery? Explain. * Rank the currents iA, iB, iC and iBat, Explain your reasoning. b. A long time after the switch is closed: * Rank the currents iA, iB, iC and iBat, Explain your reasoning. * What is the potential difference across bulb A, B, across the capacitor C and the battery? Explain. c. Summarize your results by describing the behavior of bulb A and of bulb B from just after the switch is closed until a long time later.Explanation / Answer
a. We know that voltage across capacitor do not change instantaneously.So as capacitor was initially uncharged,just after closing the switch ,voltage across capacitor willremain zero.So it will act as a short circuit shorting bulb b.
Thus VA=Vbat, VB=VC=0 .
Let the resistance of bulb A and bulb B be r(since both are identical).
thus iA=ibat=iC=Vbat/r, iB=0.
b)After long time, The voltage across capacitor becomes constant.So the current(cdv/dt) becomes zero.So it behaves as an open circuit.
Thus iA=iB=ibat=Vbat/2r, iC=0
and VA=VB=VC=Vbat/2.
So we can see that voltage across bulb A reduces to half whereas across bulb B it increases from 0 to Vbat/2.
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