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PhysicsEMI/ACL-R circuitMedium2 minQB

PhysicsMediumnumerical

In the circuit shown, $S_{1}$ is kept closed for long time. Then at $t = 0$ , Switch is thrown to $S_{2}$ position so that $S_{1}$ is open and $S_{2}$ closed. Then at $t = In 2$ , what is value of current in the inductor in ampere?

Answer:

4.50

Solution:

$t = 0$ current through inductor $= 4 a$

Time constant from lower loop is $\frac{L}{R} = \frac{2}{2} = 1$

So current as a function of time in the loop is given by $I = 4 e^{- t} + \frac{10}{2} [1 - e^{- t}] I = (4 - 5) e^{- t} + 5 I = 5 - \frac{1}{e ^{t}}$

Put $t = 1 n 2 I = 5 - \frac{1}{2} = 4.5 A$

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

⏱ 2mℹ️ Source: QB

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