A battery, a capacitor, a bulb, and a switch are in a circuit in series. Unlike

Question

A battery, a capacitor, a bulb, and a switch are in a circuit in series. Unlike most real bulbs, the resistance of the bulb in this question does not change as the current through it changes. The switch is initially open, and the capacitor is uncharged. Which statement correctly describes the bulb after the switch is closed? Assume the battery is ideal (it has no internal resistance) and the connecting wires have no resistance. 1. The bulb is bright and remains bright. 2. None of these is correct. 3. The bulb is dim and remains dim 4. At first the bulb is bright and it gets dimmer and dimmer until it goes off. 5. At first the bulb is dim and it gets brighter and brighter until the brightness levels off.

Which of the following correctly describes what happens after the switch has remained closed for a long time? 1. None of these is correct. 2. The bulb continues is permanently on. 3. The potential difference across the capacitor is steady and much smaller than E. 4. The bulb is permanently off. 5. The current in the circuit is steady.

Explanation / Answer

A)

answer : 4. At first the bulb is bright and it gets dimmer and dimmer until it goes off.

EXPLANATION : The capacitor in the circuit acts as a short circuit initially when the switch is closed. Thus the circuit has the highest current and thus it glows the highest initially. But with time the capacitor charges up and after a long time it charges up completely and shuts off the current in the circuit due to it, thus the bulb goes off.

B)

answer : 4. The bulb is permanently off.

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