Current in a coil changes from 4 ,A to zero i | vedclass

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

Question

(C) The magnitude of the induced emf in a coil is given by the formula: $|e| = L \left| \frac{dI}{dt} \right|$.Given:Initial current $I_1 = 4 \,A$.Fin

Vedclass · Accepted Answer

$2.5$

Vedclass · Answer

(C) The magnitude of the induced emf in a coil is given by the formula: $|e| = L \left| \frac{dI}{dt} ight|$.Given:Initial current $I_1 = 4 \,A$.Final current $I_2 = 0 \,A$.Change in current $dI = I_2 - I_1 = 0 - 4 = -4 \,A$.Time interval $dt = 0.1 \,s$.Induced emf $|e| = 100 \,V$.Substituting these values into the formula:$100 = L 	imes \frac{|-4|}{0.1}$.$100 = L 	imes \frac{4}{0.1}$.$100 = L 	imes 40$.$L = \frac{100}{40} = 2.5 \,H$.Therefore,the self-inductance of the coil is $2.5 \,H$.

Current in a coil changes from $4 \,A$ to zero in $0.1 \,s$ and the emf induced is $100 \,V$. The self inductance of the coil is (in $\,H$)

Similar Questions

$A$ choke coil works on the principle of

When the number of turns in a coil is made $3$ times without any change in the length of the coil,its self-inductance becomes

An air core solenoid has $1000$ turns and is one metre long. Its cross-sectional area is $10 \, cm^2$. Its self-inductance is: (in $, mH$)

The current through an inductor is given by $i = (10 + 20t) \, A$,where $t$ is in $s$. The induced $emf$ in the inductor is $4 \, V$. The total flux linked with the inductor at $t = 2 \, s$ is ...... $Wb$.

In a coil,current falls from $5\, A$ to $0\, A$ in $0.2\, s$. If an average $emf$ of $150\, V$ is induced,then the self-inductance of the coil is.....$H$

Vedclass Test Series

Exam Paper Generator

Online Exam Module