A series combination of two capacitors $C_1 = 3\, μF$ and $C_2 = 6\, μF$ is connected in parallel with a third capacitor, $C_3 = 4\, μF$. This system is connected to a 10 V battery. The total energy stored in the system is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $3 × 10^{-4} J$
Given:
$C_1 = 3\ \mu F,\ C_2 = 6\ \mu F,\ C_3 = 4\ \mu F,\ V = 10\ V$
Series combination:
$\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{3} + \frac{1}{6} = \frac{1}{2}$
$C_s = 2\ \mu F$
Parallel combination:
$C_{eq} = C_s + C_3 = 2 + 4 = 6\ \mu F$
Total energy stored:
$U = \frac{1}{2} C_{eq} V^2 = \frac{1}{2} \times 6\times10^{-6} \times (10)^2$
$U = 3\times10^{-4}\ \text{J}$
Therefore, total energy stored = $3\times10^{-4}\ \text{J}$