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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 V
Given, 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 LR circuit is,

τ=LR=20×10310=2×103 s

At any instant time t current in LR growth circuit is,

i=i0⎢ ⎢1e(tτ)⎥ ⎥

i=ER⎢ ⎢1e(tτ)⎥ ⎥

i=510⎢ ⎢1e(t2×103)⎥ ⎥

i=0.5⎢ ⎢1e(t2×103)⎥ ⎥

Differentiating the above relation with time t we get,

didt=ddt⎜ ⎜0.5⎢ ⎢1e(t2×103)⎥ ⎥⎟ ⎟

=⎢ ⎢0.5×e(t2×103)×12×103⎥ ⎥

Putting this value in (1) we get,

|Ei|=20×103×⎢ ⎢0.5×e(t2×103)×12×103⎥ ⎥

=5×e(t2×103)

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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