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A current $I= 10 \sin (100 \pi t)$ ampere is passed in first coil which induces a maximum emf of $5 \pi$ volts in second coil. The mutual inductance M is

$(a)\;10\;mH \\ (b)\;5\;mH \\(c)\;25\;mH \\(d)\;15\;mH $

1 Answer

Given $I= 10 \sin (100 \pi T)$
Comparing with $I=I_0 \sin wt$
$w= 100 \pi$
$e _{max}=M I_o w; e_{max}=3 \pi mV$
$5 \pi = M \times 10 \times 100 \pi$
$M= 5 mH$
Hence b is the correct answer.
answered Mar 21, 2014 by meena.p

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