(C) The current in an $LR$ circuit is given by $i = \frac{E}{R}(1 - e^{-t/\tau})$,where $\tau = \frac{L}{R} = \frac{1}{100} = 0.01\,s = 10\,ms$.$(a)$

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(C) The current in an $LR$ circuit is given by $i = \frac{E}{R}(1 - e^{-t/\tau})$,where $\tau = \frac{L}{R} = \frac{1}{100} = 0.01\,s = 10\,ms$.$(a)$ For the current to reach half its steady-state value $(i = \frac{E}{2R})$:$\frac{E}{2R} = \frac{E}{R}(1 - e^{-t/\tau})$$0.5 = 1 - e^{-t/\tau} \implies e^{-t/\tau} = 0.5$$t = \tau \ln 2 = 10\,ms \times 0.693 = 6.93\,ms \approx 7\,ms$.$(b)$ At $t = 15\,ms$,$t/\tau = 15/10 = 1.5$:$i = \frac{6}{100}(1 - e^{-1.5}) = 0.06(1 - 0.25) = 0.06 \times 0.75 = 0.045\,A$.The energy stored is $U = \frac{1}{2}Li^2 = \frac{1}{2} \times 1 \times (0.045)^2 = 0.5 \times 0.002025 \approx 0.001\,J = 1\,mJ$.

Class 12 Physics Electromagnetic Induction R-L D.C. Circuit

Difficult

$A$ coil of inductance $1\,H$ and resistance $100\,\Omega$ is connected to a battery of $6\,V$. Determine approximately:
$(a)$ The time elapsed before the current acquires half of its steady-state value.
$(b)$ The energy stored in the magnetic field associated with the coil at an instant $15\,ms$ after the circuit is switched on. (Given $\ln 2 = 0.693$,$e^{-3/2} = 0.25$)

JEE MAIN 2022 All JEE MAIN

A
$t = 10\,ms; U = 2\,mJ$
B
$t = 10\,ms; U = 1\,mJ$
C
$t = 7\,ms; U = 1\,mJ$
D
$t = 7\,ms; U = 2\,mJ$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Physics Questions

Chapter

Electromagnetic Induction

Subtopic

R-L D.C. Circuit

Previous Year

JEE MAIN 2022 Paper All JEE MAIN years →

An inductor $(L = 0.03 \; H)$ and a resistor $(R = 0.15 \; k\Omega)$ are connected in series to a battery of $15 \; V$ $EMF$ in the circuit shown below. The key $K_1$ has been kept closed for a long time. Then at $t = 0$,$K_1$ is opened and key $K_2$ is closed simultaneously. At $t = 1 \; ms$,the current in the circuit will be ........... $mA$. $(e^5 \cong 150)$

DifficultJEE MAIN 2015

View Solution

$A$ coil of inductance $5\,H$ is joined to a cell of $emf$ $6\,V$ through a resistance $10\,\Omega$ at time $t = 0$. The $emf$ across the coil at time $t = \ln \sqrt{2}\,s$ is:....$V$

View Solution

$A$ solenoid has an inductance of $60\, H$ and a resistance of $30\, \Omega$. If it is connected to a $100\, V$ battery,how long will it take for the current to reach $\frac{e - 1}{e} \approx 63.2\%$ of its final value?

Difficult

View Solution

Three identical resistors with resistance $R = 12\,\Omega$ and two identical inductors with self-inductance $L = 5\,mH$ are connected to an ideal battery with an emf of $12\,V$ as shown in the figure. The current through the battery long after the switch has been closed will be $......A$.

MediumJEE MAIN 2023

View Solution

An inductor with an iron core is connected in series with a resistor $R$ across an ideal $DC$ source. The circuit is in a steady state. If the iron core is removed, what happens to the current flowing through the inductor immediately after the removal of the core?

Easy

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Similar Questions

$A$ coil of inductance $5\,H$ is joined to a cell of $emf$ $6\,V$ through a resistance $10\,\Omega$ at time $t = 0$. The $emf$ across the coil at time $t = \ln \sqrt{2}\,s$ is:....$V$

$A$ solenoid has an inductance of $60\, H$ and a resistance of $30\, \Omega$. If it is connected to a $100\, V$ battery,how long will it take for the current to reach $\frac{e - 1}{e} \approx 63.2\%$ of its final value?

Three identical resistors with resistance $R = 12\,\Omega$ and two identical inductors with self-inductance $L = 5\,mH$ are connected to an ideal battery with an emf of $12\,V$ as shown in the figure. The current through the battery long after the switch has been closed will be $......A$.

An inductor with an iron core is connected in series with a resistor $R$ across an ideal $DC$ source. The circuit is in a steady state. If the iron core is removed, what happens to the current flowing through the inductor immediately after the removal of the core?

Vedclass Test Series

Exam Paper Generator

Online Exam Module