A constant voltage is applied to a series $\mathrm{R}-\mathrm{L}$ circuit by closing the switch. The voltage across inductor $(\mathrm{L}=2 \mathrm{H})$ is $20 \mathrm{~V}$ at $\mathrm{t}=0$ and drops to $5 \mathrm{~V}$ at $20 \mathrm{~ms}$. The value of $\mathrm{R}$ in $\Omega$ is:

JEE PHYSICS MCQ: MAGNETIC EFFECT OF CURRENT

1 Hard

1 of 1

Application +4 | -1

A constant voltage is applied to a series

\mathrm{R}-\mathrm{L}

circuit by closing the switch. The voltage across inductor

(\mathrm{L}=2 \mathrm{H})

20 \mathrm{~V}

\mathrm{t}=0

and drops to

5 \mathrm{~V}

20 \mathrm{~ms}

. The value of

\mathrm{R}

\Omega

is:

View answer and explanation

Correct option: C

100 \ln 4 \Omega

Option C (

100 \ln 4 \Omega

) is correct because it is the value of R calculated using the given voltage drop across the inductor in an RL circuit.

Practice All Chapter MCQs →

R

B_L

R/n

B_C

B_L / B_C

X

Y

R_1

R_2

{X}

{Y}