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Consider a current in a circuit falls from $6.0 \mathrm{~A}$ to $1.0 \mathrm{~A}$ in $0.2 \mathrm{~s}$. If an average emf of $150 \mathrm{~V}$ is induced by the circuit, then the self inductance of the circuit is
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Verified Answer
The correct answer is:
$6 \mathrm{H}$
Given, $I_1=6 \mathrm{~A}, I_2=1 \mathrm{~A}, \Delta t=0.2 \mathrm{~s}$ and $e=150 \mathrm{~V}$
Average emf induced in a inductor,
$$
e=L \frac{\left(Y_1-Y_2\right)}{\Delta t}
$$
Self inductance, $L=\frac{e \cdot \Delta t}{I_1-I_2}=\frac{150 \times 0.2}{6-1}=\frac{15 \times 2}{5}$
$$
\Rightarrow \quad L=6 \mathrm{H}
$$
Hence, the correct option is (2).
Average emf induced in a inductor,
$$
e=L \frac{\left(Y_1-Y_2\right)}{\Delta t}
$$
Self inductance, $L=\frac{e \cdot \Delta t}{I_1-I_2}=\frac{150 \times 0.2}{6-1}=\frac{15 \times 2}{5}$
$$
\Rightarrow \quad L=6 \mathrm{H}
$$
Hence, the correct option is (2).
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