In the circuit shown, the energy stored in the capacitor is $$n$$ $$\mu$$J. The value of $$n$$ is _____.

At steady state capacitor carries no current, so ignore capacitor branch while finding voltages.

The diamond becomes two parallel branches across 12 V source.

Left branch:

$$3\Omega+9\Omega=12\Omega$$

Current through left branch:

$$I_1=\frac{12}{12}=1A$$

Potential drop across $$3Ω$$:

$$V=1(3)=3V$$

So potential at left capacitor terminal (junction of $$3Ω\ and\ 9Ω$$) is

$$12-3=9V$$

Right branch:

$$4Ω+2Ω=6Ω$$

Current:

$$I_2=\frac{12}{6}=2$$

$$Dropnacrossn4Ω:V=2(4)=8V$$

Potential at right capacitor terminal:

$$12−8=4V$$

Voltage across capacitor:

$$V_C=9-4=5V$$

Capacitance:

$$C=6μF$$

Energy stored:

$$U=\frac{1}{2}CV^2$$

$$=\frac{1}{2}(6)(5^2)\mu J$$

$$=75μJ$$