In the circuit shown below, the key K is closed at t = 0. The current through the battery is (a) (VR₁R₂)/(√(R₁²+R₂²)) at t = 0 and V/R₂ at t = ∞ (b) V/R₂ at t = 0 and (V(R₁+R₂))/(R₁R₂) at t = ∞ (c) V/R₂ at t = 0 and (VR₁R₂)/(√(R₁²+R₂²)) at t = ∞ (d) (V(R₁+R₂))/(R₁R₂) at t = 0 and V/R₂ at t = ∞

Answer: (b). At t = 0 the inductor carries no current, so only R₂ draws current: i = V/R₂. At t = ∞ the inductor is a plain wire, so R₁ and R₂ are both directly across V. i = V/R₁ + V/R₂ = (V(R₁+R₂))/(R₁R₂) So the answer is (b).

In the circuit shown below, the key K is closed at t = 0. The current… | JEE Main 2010 (AIEEE)

Q29JEE Main 2010 · AIEEEMedium

Inductors & DC Transients (Mixed)

In the circuit shown below, the key $K$ is closed at $t = 0.$ The current through the battery is

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Answer: (b)

At $t = 0$ the inductor carries no current, so only $R_2$ draws current: $i = \frac{V}{R_2}$ .
At $t = \infty$ the inductor is a plain wire, so $R_1$ and $R_2$ are both directly across $V$ .
$i = \frac{V}{R_1} + \frac{V}{R_2} = \frac{V(R_1+R_2)}{R_1R_2}$
So the answer is (b).

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