The current (i) at time t = 0 and t = ∞ respectively for the given circuit is: (a) 18E/55, 5E/18 (b) 5E/18, 18E/55 (c) 5E/18, 10E/33 (d) 10E/33, 5E/18

Answer: (c). At t = 0 the inductor carries no current, so the circuit is two series branches across E: 5 + 1 = 6 Ω and 5 + 4 = 9 Ω. R₀ = (6 × 9)/(6 + 9) = 3.6 Ω ⇒ i₀ = E/3.6 = 5E/18 At t = ∞ the inductor shorts the two side nodes: (5∥5) + (1∥4) = 2.5 + 0.8 = 3.3 Ω i∞ = E/3.3 = 10E/33 So the answer is (c).

The current (i) at time t = 0 and t = ∞ respectively for the given… | JEE Main 2021 (25 Feb, Shift 1)

Q15JEE Main 2021 · 25 FEB · SHIFT 1Difficult

Inductors & DC Transients (Mixed)

The current (i) at time t = 0 and t = $\infty$ respectively for the given circuit is:

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Answer: (c)

At $t = 0$ the inductor carries no current, so the circuit is two series branches across E: $5 + 1 = 6\ \Omega$ and $5 + 4 = 9\ \Omega$ .
$R_0 = \frac{6 \times 9}{6 + 9} = 3.6\ \Omega \Rightarrow i_0 = \frac{E}{3.6} = \frac{5E}{18}$
At $t = \infty$ the inductor shorts the two side nodes: $(5\parallel5) + (1\parallel4) = 2.5 + 0.8 = 3.3\ \Omega$
$i_\infty = \frac{E}{3.3} = \frac{10E}{33}$
So the answer is (c).

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