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Question
An inductor of inductance 5.0 H, having a negligible resistance, is connected in series with a 100 Ω resistor and a battery of emf 2.0 V. Find the potential difference across the resistor 20 ms after the circuit is switched on.
Solution
Given:-
Self-inductance, L = 5.0 H
Resistance, R = 100 Ω
Emf of the battery = 2.0 V
At time, t = 20 ms ....(after switching on the circuit)
t = 20 ms = 20 × 10−3 s = 2 × 10−2 s
The steady-state current in the circuit is given by
\[i_0 = \frac{2}{100}\]
The time constant is given by
\[\tau = \frac{L}{R} = \frac{5}{100}\]
The current at time t is given by
i = i0(1 − e−t/τ)
\[i = \frac{2}{100}\left( 1 - e^\left( \frac{- 2 \times {10}^{- 2} \times 100}{5} \right) \right)\]
\[ = \frac{2}{100}(1 - e^{- 2/5} )\]
\[ = \frac{2}{100}(1 - 0 . 670)\]
= 0.00659 = 0.0066
Now,
V = iR = 0.0066 × 100
= 0.66 V
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