2. Two identical bulbs and a capacitor ideal battery as shown. The capacitor is
ID: 3280340 • Letter: 2
Question
2. Two identical bulbs and a capacitor ideal battery as shown. The capacitor is initially uncharged are connected to arn Bulb A Battery a. Just after the switch is closed: bat Switch what is the potential difference across bulb A. across bulb B, across the capacitor C, and across the battery? Explain. Bulb B Capacitor C rank the currents i, ig, ic, and i Explain your reasoning. b. A long time after the switch is closed: rank the currents i, ig, ic, and ir Explain your reasoning. what is the potential difference across bulb A, bulb B, the capacitor C, and the battery? Explain. 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. c.Explanation / Answer
Considering the given circuit,
a) Initially the capacitor acts like a wire and shorts out bulb B so all essentially all we have is a battery connected across bulb A. By Kirchoff's loop rule then, V(A) = V(bat). Bulb B has no current running through it (it is shorted) so V(B) = 0. The capacitor acts like a wire so V(C) = 0. Ideal batteries by definition supply a constant voltage so V(bat) = whatever voltage it normally is.
i(A) = i(bat) = i(C) > i(B) = 0 Bulb B is shorted so no current flows through it. By the node rule, the current now has only one way to flow so the rest of the currents are all equal.
b) After a long time, the capacitor acts like an object with infinite resistance. So no current flows through it. There is now only one loop for the current. Thus, i(A) = i(B) = i(bat) > i(C)=0. The battery is ideal so it has the same constant voltage across it as before. The voltage across A has decreased from before and is now equal to the voltage across B because they are identical and have the same current through them. The capacitor has the same voltage across it as B because they are in parallel.
c) Bulb A starts out pretty bright then decreases to half brightness when the switch is closed. Bulb B starts out unlit then has the same brightness as A when the switch is closed. See above for an explanation.
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