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In the circuit shown, the energy stored in the capacitor is n \muJ. The value of n is

Option: 1

75


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

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\begin{aligned} & \mathrm{I}_1=\frac{12}{3+9}=1 \mathrm{~A} \\ & \mathrm{I}_2=\frac{12}{4+2}=2 \mathrm{~A} \\ & \mathrm{~V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{C}}=3 \mathrm{I}_1=3 \mathrm{~V} \\ & \mathrm{~V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{D}}=2 \times 4=8 \mathrm{~V} \\ & \text { So, } \mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{D}}=5 \mathrm{~V} \\ & \mathrm{U}=\frac{1}{2} \mathrm{CV}^2=\frac{1}{2} \times 6 \times 5^2=75 \mu \mathrm{J} \end{aligned}

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Gautam harsolia

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