Question

In the figure given below, Find the
total energy stored in the network.

Answer 1

Given circuit,

$$(C_1 = C_5 = 4 \mu F, C_2 = C_3 = C_4 = 2 \mu F)$$

Ref. image

From about circuit diagram $$C_1 $$ & $$C_2$$ and $$C_4$$ & $$C_5$$ are shorted with wire. So total potential drop to move from $$C_1 $$ to $$C_2 $$ and $$C_4 $$ to $$C_5$$ will be zero

$$\therefore C_{eq} = C_3 = 2 \mu F$$

Charge supply by battery

$$q = C_{eq} \times V$$

$$= 2 \times 10^{-6} \times 5$$

$$q = 10 \times 10^{-6}$$

$$q = 10 \mu C$$

total energy stored

$$U = \dfrac{1}{2} (q v) = \dfrac{1}{2} \times 10 \times 5 = 25 \mu J$$