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Standard XII

Physics

Lenz's Law

Two circuits ...

Question

Two circuits have mutual inductance of 0.09 H. Average e.m.f. induced in the secondary by a change of current from 0 to 20 A in 0.006 s in primary will be

120 V

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200 V

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180 V

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300 V

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Solution

The correct option is D
300 V

Given $Δ I = 20 A, Δ t = .006 s$
$∴ \frac{Δ I}{Δ t} = \frac{20}{.006}$
$∴ E = .09 \times \frac{20}{.006} = \frac{180}{.6} = 300 V$

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Two circuits have mutual inductance of 0.09 H. Average e.m.f. induced in the secondary by a change of current from 0 to 20 A in 0.006 s in primary will be

The correct option is D 300 V Given ΔI=20A,Δt=.006 s ∴ΔIΔt=20.006 ∴E=.09×20.006=180.6=300 V

The correct option is D
300 V

Given $Δ I = 20 A, Δ t = .006 s$
$∴ \frac{Δ I}{Δ t} = \frac{20}{.006}$
$∴ E = .09 \times \frac{20}{.006} = \frac{180}{.6} = 300 V$