Question

A fully charged capacitor connected in series with a resistor stores 64J of energy. How much energy remains when its charge has decreased to one quarter of its original value?

Answer 1

Question 27298

The energy of the capacitor $W = \frac{CU^2}{2} = \frac{qU}{2}$ .

Also, $W_{0} = \frac{q_{0}U}{2} = 64J$ .

If the change in charge is $q \to \frac{q}{4}$ , then new value of energy of capacitor is

$$W _ {1} = \frac {q _ {0} U}{8} = \frac {W _ {0}}{4} = 16 J.$$

Hence, $16J$ of energy remains when charge has decreased to quarter of its original value.