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

At t = 0 , current through inductor is zero, hence R e q = ( 5 + 1 ) ∥ ( 5 + 4 ) = 5 18 i 1 = 18/5 E = 18 5 E At t = ∞ , inductor becomes a simple wire and now the circuit will be as shown in figure hence R e q = ( 5 ∥ 5 ) + ( 4 ∥ 1 ) = 10 33 ; ( ∥⇒ parallel) i 2 = 33/10 E = 33 10 E

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PhysicsEMI/ACL-R circuitHard2 minQB

PhysicsHardsingle choice

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

Options:

Answer:

Solution:

At $t = 0$ , current through inductor is zero,

hence $R_{e q} = (5 + 1) ∥ (5 + 4) = \frac{18}{5}$

$i_{1} = \frac{E}{18/5} = \frac{5 E}{18}$

At $t = \infty,$ inductor becomes a simple wire and now the circuit will be as shown in figure

hence $R_{e q} = (5 ∥ 5) + (4 ∥ 1) = \frac{33}{10};$ ( $∥\Rightarrow$ parallel)

$i_{2} = \frac{E}{33/10} = \frac{10 E}{33}$

Stream:JEESubject:PhysicsTopic:EMI/ACSubtopic:L-R circuit

⏱ 2mℹ️ Source: QB

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