Two coils A and B have 180 and 360 turns respectively. The current of… | MHT CET 2026 (18 Apr, Shift 2)

Q2MHT CET 2026 · 18 APR · SHIFT 2Easy

Self-Inductance & Energy

Two coils A and B have 180 and 360 turns respectively. The current of 1A flows through both the coils. Due to current of 1A in coil A, flux per turn of $0.8 \times 10^{-3}$ Wb is linked with coil A. Due to current of 1A in coil B, flux per turn of $1 \times 10^{-3}$ Wb is linked with coil B. The self inductance of coil A is $L_A$ and the self inductance of coil B is $L_B$ . The ratio $L_A$ to $L_B$ is

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Answer: (b)

Self-inductance $L = \frac{N\phi}{I}$ (flux per turn $\phi$ ).
$L_A = \frac{180 \times 0.8\times10^{-3}}{1} = 0.144\text{ H}$ , $L_B = \frac{360 \times 1\times10^{-3}}{1} = 0.36\text{ H}$
$\frac{L_A}{L_B} = \frac{0.144}{0.36} = \frac{2}{5}$
So the answer is (b) $\frac{2}{5}$ .

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