An ideal coil of 10 H is connected in series with a resistance of 5 Ω and a battery of 5 V · 2 second after the connection is made, the current flowing in ampere in the circuit is (a) (1 - e⁻¹) (b) (1 - e) (c) e (d) e⁻¹

Answer: (a). I = E/R(1 - e-Rt/L), with E/R = 1 A Rt/L = (5×2)/(10) = 1 I = (1 - e⁻¹) A So the answer is (a).

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Combination of Inductors, LR & LC Circuits

An ideal coil of $10\text{ H}$ is connected in series with a resistance of $5\,\Omega$ and a battery of $5\text{ V} \cdot 2$ second after the connection is made, the current flowing in ampere in the circuit is

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Answer: (a)

$I = \frac{E}{R}(1 - e^{-Rt/L})$ , with $\frac{E}{R} = 1\text{ A}$
$\frac{Rt}{L} = \frac{5\times2}{10} = 1$
$I = (1 - e^{-1})\text{ A}$
So the answer is (a).

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