In the circuit shown, the energy stored in the capacitor is $n\,\mu J$. The value of $n$ is ..............
JEE MAIN 2023, Diffcult
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$I _1=\frac{12}{3+9}=1\,A$
$I _2=\frac{12}{4+2}=2\,A$
$V_A-V_C=3 I_1=3\,V$
$V_A-V_D=2 \times 4=8\,V$
Subtracting eq.$(1)$ from eq.$(2)$
$V _{ C }- V _{ D }=5 V \Rightarrow V =5\,V$
$U =\frac{1}{2} CV ^2=\frac{1}{2} \times 6 \times 5^2=75\,\mu J$
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