JEE MAIN - Physics (2024 - 29th January Morning Shift - No. 27)
A $$16 \Omega$$ wire is bend to form a square loop. A $$9 \mathrm{~V}$$ battery with internal resistance $$1 \Omega$$ is connected across one of its sides. If a $$4 \mu F$$ capacitor is connected across one of its diagonals, the energy stored by the capacitor will be $$\frac{x}{2} \mu J$$, where $$x=$$ _________
Answer
81
Explanation
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$$\begin{aligned} & I=\frac{V}{R_{\text {eq }}} I=\frac{V}{R_{\text {eq }}}=\frac{9}{1+\frac{12 \times 4}{12+4}}=\frac{9}{4} \\ & I_1=\frac{9}{4} \times \frac{4}{16}=\frac{9}{16} \\ & V_A-V_B=I_1 \times 8=\frac{9}{16} \times 8=\frac{9}{2} V \\ & \therefore U=\frac{1}{2} \times 4 \times \frac{81}{4} \mu J \\ & \therefore U=\frac{81}{2} \mu J \\ & \therefore x=81 \end{aligned}$$
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