Question

The circuit shown in Figure is in steady state.
i. Find the energy stored in the capacitors shown in Figure.
ii. Find the rate at which battery supplies energy.

Answer 1

Since, in steady state, no current flows through the capacitors, the current through $1\Omega$ resistor becomes zero. Current through resistors and charge on capacitors will be as shown in Figure.

Applying $KVL$ on mesh $MACDA$, we get

$2I+3I+3I+2I-10=0$ or $I=1A$

Mesh $MABM:10-2I-\frac{q_1}{2\times {10}^{-6}}=0$ or $q_1=16\mu C$

Mesh $MBDM:-\frac{q_2}{2\times {10}^{-6}}-2I=0$ or $q_2=-4\mu C$

Mesh $MDCNM:2I+3I-\frac{q_3}{(1\times {10}^{-6})}$ or $q_3=5\mu C$

Energy stored in capacitors is

$U=\sum \frac{q^2}{2C}=\frac{q_1^2}{2\times (2\times {10}^{-6})}+\frac{q_2^2}{2\times (2\times {10}^{-6})}+\frac{q_3^2}{2\times (1\times {10}^{-6})}$

$=80.5\times {10}^{-6}J$

Rate of supply of energy is $P=EI=10\times 1=10W$