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The capacitor is initially discharged, and the battery is at a positive voltage. Which correctly describes what happens after the switch has remained closed for a long time? Note: The battery is ideal (no internal resistance). The wires are ideal (no connecting resistance). The resistance of the bulb does not change as current flows through it. The switch is ideal; it has no resistance when closed, and infinite resistance when open. The capacitor is ideal (no internal resistance or inductance).

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klefenz

Answer:

The lamp will remain off.

Explanation:

I assume this is a simple series circuit with a battery, a switch, an incandescent lamp and a capacitor.

Since the circuit is of continuous current (because it uses a battery), after the switch has remained closed for a long time the capacitor will be fully charged. Being fully charged, no continuous current will flow through the capacitor, and since it is in series with the rest of the circuit there will be no current anywhere, so the lamp will remain off.

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