Question

A capacitor of capacity \(10 \, \mu F\) is charged with a source of \(2\) volt. Calculate (i) the energy obtained from the source and (ii) the energy stored in the capacitor.

Hint:

In capacitors, only half of the energy supplied by the source is stored; the rest is dissipated in the circuit.

Answer 1

Step 1: The total energy supplied by the source is calculated using: \[ \text{Energy} = CV^2 \] \[ = (10 \times 10^{-6}) (2)^2 \] \[ = 4 \times 10^{-5} \, \text{J} \] Step 2: The energy stored in the capacitor is: \[ \text{Stored Energy} = \frac{1}{2} CV^2 \] \[ = \frac{1}{2} (10 \times 10^{-6}) (2)^2 \] \[ = 2 \times 10^{-5} \, \text{J} \] \[ \therefore \text{The correct answers are } 4 \times 10^{-5} \, \text{J} \text{ and } 2 \times 10^{-5} \, \text{J}. \] \[ \boxed{4 \times 10^{-5} \, \text{J}, \, 2 \times 10^{-5} \, \text{J}} \]