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A current \mathrm{I=10 \sin (100 \pi t)} amp. is passed in first coil, which induces a maximum e.m.f of \mathrm{5 \pi} volt in second coil. The mutual inductance between the coils is -

Option: 1

10 mH


Option: 2

15 mH


Option: 3

25 mH


Option: 4

5 mH


Answers (1)

best_answer

\mathrm{\text { Let } \mathrm{I}=\mathrm{I}_0 \sin \omega \mathrm{t} \text {, }}

\mathrm{\text { where } I_0=10, \omega=100 \pi}

\mathrm{\text { then } \varepsilon=M \frac{d I}{d t}}

\mathrm{\begin{aligned} & \quad=M \frac{d}{d t} I_0 \sin \omega t \\ & =M I_0 \omega \cos \omega t \\ & \therefore \varepsilon_{\max }=M I_0 \omega \\ & 5 \pi=M \times 10 \times 100 \pi \\ & M=5 \mathrm{mH} \end{aligned}}

 

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Rishabh

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