A 12,V battery connected to a coil of resista | vedclass

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

Question

(D) The initial current $I$ in the circuit is given by Ohm's law: $I = \frac{V}{R} = \frac{12\,V}{6\,\Omega} = 2\,A$.When the switch is opened,the cur

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(D) The initial current $I$ in the circuit is given by Ohm's law: $I = \frac{V}{R} = \frac{12\,V}{6\,\Omega} = 2\,A$.When the switch is opened,the current drops from $2\,A$ to $0\,A$ in a time interval $\Delta t = 1\,ms = 10^{-3}\,s$.The magnitude of the induced emf is given by the formula: $|\varepsilon| = L \left| \frac{\Delta I}{\Delta t} ight|$.Substituting the given values: $20 = L 	imes \frac{2 - 0}{10^{-3}}$.$20 = L 	imes \frac{2}{10^{-3}}$.$20 = L 	imes 2000$.$L = \frac{20}{2000} = 0.01\,H$.Converting to millihenry $(mH)$: $L = 0.01 	imes 1000\,mH = 10\,mH$.

$A$ $12\,V$ battery connected to a coil of resistance $6\,\Omega$ through a switch,drives a constant current in the circuit. The switch is opened in $1\,ms$. The emf induced across the coil is $20\,V$. The inductance of the coil is ........... $mH$.

Similar Questions

$A$ coil having an inductance of $0.5\, H$ carries a current which is uniformly varying from zero to $10\, A$ in $2\, s$. The $e.m.f.$ (in volts) generated in the coil is

The current through a coil varies according to the following graph. Plot the graph for induced $emf$ versus time for the coil.

The current in a coil decreases from $1 \, A$ to $0.2 \, A$ in $10 \, s$. Calculate the coefficient of self-inductance if the induced $e.m.f.$ is $0.4 \, V$. (in $, H$)

$A$ graph of magnetic flux $(\Phi)$ versus current $(I)$ is shown for four inductors $A, B, C, D$. The smallest value of self-inductance is for inductor:

The self-inductance of a circuit is numerically equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module