JEE Mains · Physics · STD 12 - Current electricity
- JEE Mains 2023
- Hard
- Numerical
In the circuit shown, the energy stored in the capacitor is \(n\,\mu J\). The value of \(n\) is ..............

Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(75\)
Step-by-step Solution
Detailed explanation
\(I _1=\frac{12}{3+9}=1\,A\)
\(I _2=\frac{12}{4+2}=2\,A\)
\(V_A-V_C=3 I_1=3\,V\)
\(V_A-V_D=2 \times 4=8\,V\)
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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