JEE MAIN - Physics (2023 - 13th April Evening Shift - No. 26)
In the circuit shown, the energy stored in the capacitor is $$n ~\mu \mathrm{J}$$. The value of $$n$$ is __________
_13th_April_Evening_Shift_en_26_1.png)
Answer
75
Explanation
$$
\begin{aligned}
& \mathrm{I}_1=\frac{12}{3+9}=1 \mathrm{~A} \\\\
& \mathrm{I}_2=\frac{12}{4+2}=2 \mathrm{~A}
\end{aligned}
$$
_13th_April_Evening_Shift_en_26_2.png)
$$ \begin{aligned} & \mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{C}}=3 \mathrm{I}_1=3 \mathrm{~V} \\\\ & \mathrm{~V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{D}}=2 \times 4=8 \mathrm{~V} \end{aligned} $$
Subtracting eq. (1) from eq. (2)
$$ \begin{aligned} & \mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{D}}=5 \mathrm{~V} \Rightarrow \mathrm{V}=5 \mathrm{~V} \\\\ & \mathrm{U}=\frac{1}{2} \mathrm{CV}^2=\frac{1}{2} \times 6 \times 5^2=75 \mu \mathrm{J} \end{aligned} $$
$$ \begin{aligned} & \mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{C}}=3 \mathrm{I}_1=3 \mathrm{~V} \\\\ & \mathrm{~V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{D}}=2 \times 4=8 \mathrm{~V} \end{aligned} $$
Subtracting eq. (1) from eq. (2)
$$ \begin{aligned} & \mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{D}}=5 \mathrm{~V} \Rightarrow \mathrm{V}=5 \mathrm{~V} \\\\ & \mathrm{U}=\frac{1}{2} \mathrm{CV}^2=\frac{1}{2} \times 6 \times 5^2=75 \mu \mathrm{J} \end{aligned} $$
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