If a current of 3 amperes flowing in the primary coil is reduced to zero in 0.001second, then the induced e.m.f in the secondary coil is 15000 volts. The mutual inductance between the two coils is

0.5 H

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5 H

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1.5 H

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10 H

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Solution

The correct option is B
5 H

$ϵ = - M \frac{Δ I}{Δ t} 15 \times 10^{3} = - M \times \frac{(0 - 3)}{0.001} \Rightarrow M = 5 H$

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If a current of 3 amperes flowing in the primary coil is reduced to zero in 0.001second, then the induced e.m.f in the secondary coil is 15000 volts. The mutual inductance between the two coils is

The correct option is B 5 H ϵ=−MΔIΔt15×103=−M×(0−3)0.001⇒M=5H

The correct option is B
5 H

$ϵ = - M \frac{Δ I}{Δ t} 15 \times 10^{3} = - M \times \frac{(0 - 3)}{0.001} \Rightarrow M = 5 H$