# Thermal heat from a Capacitor

## Homework Statement

A resistor, a capacitor, a battery and a switch are connected to a circuit. The capacitor will be charged from the empty state to the full state.

During the period in which the capacitor is charged, how much thermal heat is emitted from the resistor?

$$C = 2 \mu F$$
$$R = 2 \Omega$$
$$E = 6 V$$

## Homework Equations

$$U = RI$$
$$Q = CV$$

## The Attempt at a Solution

The amount of charge stored in the capacitor is: 6 V * 2 $$\mu F$$ = 12 $$\mu C$$.

The amount of work done by the battery is 12 $$\mu C$$* 6 V *1/2 = 36 $$\mu J$$, during charging.

The amout of heat is (wrongly) 3/4 * 36 $$\mu J$$.

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Andrew Mason
Homework Helper

## Homework Statement

A resistor, a capacitor, a battery and a switch are connected to a circuit. The capacitor will be charged from the empty state to the full state.

During the period in which the capacitor is charged, how much thermal heat is emitted from the resistor?

$$C = 2 \mu F$$
$$R = 2 \Omega$$
$$E = 6 V$$

## Homework Equations

$$U = RI$$
$$Q = CV$$

## The Attempt at a Solution

The amount of charge stored in the capacitor is: 6 V * 2 $$\mu F$$ = 12 $$\mu C$$.

The amount of work done by the battery is 12 $$\mu C$$* 6 V *1/2 = 36 $$\mu J$$, during charging.

The amout of heat is (wrongly) 3/4 * 36 $$\mu J$$.
What is the energy stored in the capacitor in terms of C and V?

Assume the energy stored in the capacitor is transformed into heat in the resistor when the capacitor discharges. How does this relate to the reverse when it charges?

AM

What is the energy stored in the capacitor in terms of C and V?

Assume the energy stored in the capacitor is transformed into heat in the resistor when the capacitor discharges. How does this relate to the reverse when it charges?

AM
The energy (=QV) is 12 $$\mu C$$ * 6 V = 66 $$\mu J$$. But somethig must be wrong, as I have:
Horse said:
The amount of work done by the battery is 12 $$\mu C$$* 6 V *1/2 = 36 $$\mu J$$, during charging.
Perhaps, it is 12 $$\mu C$$* 6 V = 72 $$\mu J$$. ??? Dunno why ???

If we suppose the logic right, we get a result for the energy emitted:

72 $$\mu J$$ - 66 $$\mu J$$ = 6 $$\mu J$$

The amount of the thermal heat emitted is the difference between the energy emitted during discharging and the work done by the battery.

After-thought
NB. If I am right, the unit check shows that my calculations are wrong. E != CQ, E = C*Q^2. Now, I must be mistaken.

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