An ideal coil of 10 H is connected in series with a resistance of $5 \Omega$ and a battery of 5 V. 2 second after the connection is made the current flowing in amperes in the circuit is

JEE Main · 2007 · 29 Apr 2007 · Offline · Electromagnetic Induction

An ideal coil of 10 H is connected in series with a resistance of

5Ω

and a battery of 5 V. 2 second after the connection is made the current flowing in amperes in the circuit is

(A) $(1 - e)$
(B)e
(C) $e^{- 1}$
(D) $(1 - e^{- 1})$ ✓ Correct

Answer: (D)

Solution

In this

L R

circuit the current grows as

i = \frac{V}{R} (1 - e^{- R t / L})

. Here the steady current is

\frac{V}{R} = \frac{5}{5} = 1

A and the time constant is

\frac{L}{R} = \frac{10}{5} = 2

s. Two seconds after the connection is made exactly one time constant has elapsed, so

i = 1 \cdot (1 - e^{- 1}) = 1 - e^{- 1}

amperes.

JEE Main · 2007 · 29 Apr 2007 · Offline · Single correct option

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