Question

# An inductor-coil of inductance 20 mH having resistance of 10 Ω is joined to an ideal battery of emf 5 V. Find the emf induced across the inductor at t=0 s.

A
5 V
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B
2.5 V
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C
1 V
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D
0.5 V
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Solution

## The correct option is A 5 VGiven, L=20 mH ; R=10 Ω ; E=5 V The induced emf across an inductor is, Ei=−L(didt) |Ei|=L(didt) ....(1) The time constant of the given L−R circuit is, τ=LR=20×10−310=2×10−3 s At any instant time t current in L−R growth circuit is, i=i0⎡⎢ ⎢⎣1−e(−tτ)⎤⎥ ⎥⎦ i=ER⎡⎢ ⎢⎣1−e(−tτ)⎤⎥ ⎥⎦ i=510⎡⎢ ⎢⎣1−e(−t2×10−3)⎤⎥ ⎥⎦ i=0.5⎡⎢ ⎢⎣1−e(−t2×10−3)⎤⎥ ⎥⎦ Differentiating the above relation with time t we get, didt=ddt⎛⎜ ⎜⎝0.5⎡⎢ ⎢⎣1−e(−t2×10−3)⎤⎥ ⎥⎦⎞⎟ ⎟⎠ =⎡⎢ ⎢⎣0.5×−e(−t2×10−3)×−12×10−3⎤⎥ ⎥⎦ Putting this value in (1) we get, |Ei|=20×10−3×⎡⎢ ⎢⎣0.5×−e(−t2×10−3)×−12×10−3⎤⎥ ⎥⎦ =5×−e(−t2×10−3) At time t=0 s |Ei|=5 V <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (A) is the correct answer. Alternate solution: At t=0, the inductor will behave like an infinite resistance. Hence, E will be 5 V as no current passes through the circuit.

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