Question

You may have noticed that while discharging a capacitor through a light bulb, the light glows just about as brightly, and for just about as long, as it does while charging the same capacitor through the same bulb. Let $\mathit{E}$stand for the energy emitted by the light bulb (as light and heat) in the discharging phase, from just before the bulb is connected to the capacitor until the time when there is essentially no more current. In terms of $\mathbf{+}\mathit{E}$or $\mathbf{-}\mathit{E}$, what was the energy change of the battery, capacitor, bulb, and surroundings during the charging phase, and during the discharging phase? One answer is already given in the following table:

It is somewhat surprising that we can get this much information out of one simple observation.

Short answer

During discharging, the energy of the battery is zero.

During charging, the energy of the battery is $-2E$.

Step-by-step solution

A concept:

The energy of the bulb during discharge of the capacitor is zero because whatever energy gained by the bulb is given in the form of light and heat to the surrounding.

The energy of the bulb during charging of the capacitor is zero because whatever energy gained by the bulb is given in the form of light and heat to the surrounding.

The energy change of the battery, capacitor, bulb, and surroundings during the charging phase, and during the discharging phase:

The energy emitted by the bulb during discharging of the capacitor as light and heat to the surrounding is $E$.

During discharge of capacitor energy of the capacitor is lost therefore energy of capacitor during discharge of the capacitor is $-E$.

The total energy of the circuit is zero in a closed path. Therefore the energy of the battery is the sum of the energy of the bulb, energy losses to the surroundings, and energy of the capacitor.

The energy of the battery is zero.

The energy emitted by the bulb during charging of the capacitor as light and heat to the surrounding is $E$.

During the charging of the capacitor energy of the capacitor is increased therefore the energy of the capacitor during discharge of the capacitor is $E$.

The total energy of the circuit is zero in a closed path. Therefore the energy of the battery is a negative sum of the energy of the bulb, energy losses to the surroundings, and energy of the capacitor.

The energy of the battery is $-2E$.

The table:

The final answer is given in the below table.

Draw the circuit diagram as below.