SATHEE: Electromagnetic Induction and Alternating Current 4 Question 4

Electromagnetic Induction and Alternating Current 4 Question 4

####4. In the circuit shown,

The switch $S_{1}$ is closed at time $t = 0$ and the switch $S_{2}$ is kept open. At some later time $(t_{0})$ , the switch $S_{1}$ is opened and $S_{2}$ is closed. The behaviour of the current $I$ as a function of time ’ $t$ ’ is given by

(Main 2019, 11 Jan I)

(a)

(b)

(c)

(d)

Show Answer

Answer:

Correct Answer: 4. (b)

Solution:

Initially in the given $R L$ circuit with a source, when $S_{1}$ is closed and $S_{2}$ is open at $t \leq t_{0}$ .

$I_{1} = \frac{V}{R} 1 - \exp \frac{- R}{L} t$

In this case, inductor $L$ is charging.

When switch $S_{2}$ is closed and $S_{1}$ is open (after $t > t_{0}$ ), the inductor will be discharged through resistor.

In this case $(t > t_{0})$ ,

$I_{2} = \frac{V}{R} \exp - \frac{R}{L} (t - t_{0})$

Thus, the variation of $I$ with $t$ approximately is shown below

Electromagnetic Induction and Alternating Current 4 Question 4

Answer:

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Electromagnetic Induction and Alternating Current 4 Question 4

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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