JEE Main · 2009 · 12 Apr 2009 · Offline · Electromagnetic Induction

An inductor of inductance L = 400 mH and resistors of resistances

R_{1} = 2 Ω

and

R_{2} = 2 Ω

are connected to a battery of emf 12 V as shown in the figure. The internal resistance of the battery is negligible. The switch S is closed at t = 0. The potential drop across L as a function of time is

(A) $6 e^{- 5 t} V$
(B) $\frac{12}{t} e^{- 3 t} V$
(C) $6 (1 - e^{- t /0.2}) V$
(D) $12 e^{- 5 t} V$ ✓ Correct

Answer: (D)

Solution

After the switch closes the current through

R_{2}

builds up as

I_{2} = I_{0} (1 - e^{- t / τ})

, with the final value

I_{0} = \frac{E}{R _{2}} = \frac{12}{2} = 6

A and the time constant

τ = \frac{L}{R _{2}} = \frac{400 \times 1 0 ^{- 3}}{2} = 0.2

s. The voltage across the inductor is what is left of the emf after the resistor drop,

V_{L} = E - R_{2} I_{2} = 12 - 12 (1 - e^{- t /0.2})

, which simplifies to

12 e^{- 5 t}

JEE Main · 2009 · 12 Apr 2009 · Offline · Single correct option

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JEE Main · 2009 · 12 Apr 2009 · Offline · Electromagnetic Induction

Solution

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