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A circuit containing a two position switch S is shown in figur

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A circuit containing a two position switch $S$ is shown in figure.
(a) The switch $S$ is in position 1. Find the potential difference $V_{A} - V_{B}$ and the rate of production of joule heat in $R_{1}$ .
(b) If now the switch $S$ is put in position 2 at $t = 0$ . Find
(i) steady current in $R_{4}$ and
(ii) the time when current in $R_{4}$ is half the steady value. Also, calculate the energy stored in the inductor $L$ at that time.

Name: Video solution 1: A circuit containing a two position switch S is shown in figure.(a) The switch S is in position 1. Find the potential difference VA−VB and the rate of production of joule heat in R1.(b) If now the switch S is put in position 2 at t=0. Find(i) steady current in R4 and(ii) the time when current in R4 is half the steady value. Also, calculate the energy stored in the inductor L at that time.
Uploaded: 2023-07-01T12:18:08.000Z
Duration: 11 min

Views: 5,715 students

Updated on: Jul 1, 2023

Exam:

JEE Advanced 1991

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A circuit containing a two position switch

S

is shown in figure.
(a) The switch

S

is in position 1. Find the potential difference

V_{A} - V_{B}

and the rate of production of joule heat in

R_{1}

.
(b) If now the switch

S

is put in position 2 at

t = 0

. Find
(i) steady current in

R_{4}

and
(ii) the time when current in

R_{4}

is half the steady value. Also, calculate the energy stored in the inductor

L

at that time.

15 mins ago

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Text solutionVerified

(a) In steady state, no current will flow through capacitor. Applying Kirchhoff's second law in loop 1
Solution Image

- 2 i_{2} + 2 (i_{1} - i_{2}) + 12 = 0 ∴ 2 i_{1} - 4 i_{2} = - 12 or i_{1} - 2 i_{2} = - 6

Applying Kirchhoff's second law in loop 2

∴ 4 i_{1} - 2 i_{2} - 12 - 2 (i_{1} - i_{2}) + 3 - 2 i_{1} = - 9 = 0

Solving Eqs. (i) and (ii), we get

i_{2} = 2.5 A and i_{1} = - 1 A

Now, V_{A} + 3 - 2 i_{1} = V_{B} or V_{A} - V_{B} = 2 i_{1} - 3 = 2 (- 1) - 3 = - 5 V P_{R_{1}} = (i_{1} - i_{2})^{2} R_{1} = (- 1 - 2.5)^{2} (2) = 24.5 W

(b) In position 2 : Circuit is as under
Solution Image

(i) Steady state current in

R_{4}

i_{0} = \frac{3}{3 + 2} = 0.6 A

(ii) Time when current in

R_{4}

is half the steady value

t U = t_{1/2} = τ_{L} (ln 2) = \frac{L}{R} ln (2) = \frac{( 10 \times 1 0 ^{- 3} )}{5} ln (2) = 1.386 \times 1 0^{- 3} s = \frac{1}{2} L i^{2} = \frac{1}{2} (10 \times 1 0^{- 3}) (0.3)^{2} = 4.5 \times 1 0^{- 4} J

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Question Text	A circuit containing a two position switch $S$ is shown in figure. (a) The switch $S$ is in position 1. Find the potential difference $V_{A} - V_{B}$ and the rate of production of joule heat in $R_{1}$ . (b) If now the switch $S$ is put in position 2 at $t = 0$ . Find (i) steady current in $R_{4}$ and (ii) the time when current in $R_{4}$ is half the steady value. Also, calculate the energy stored in the inductor $L$ at that time.
Updated On	Jul 1, 2023
Topic	Alternating Current
Subject	Physics
Class	Class 12
Answer Type	Text solution:1 Video solution: 2
Upvotes	262
Avg. Video Duration	8 min

Text solutionVerified

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