The current in an inductor of self-inductance | vedclass

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

Question

(A) Given:Self-inductance of the inductor,$L = 40 	ext{ mH} = 40 	imes 10^{-3} 	ext{ H}$.Initial current,$I_1 = 2 	ext{ A}$.Final current,$I_2 = 1

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(A) Given:Self-inductance of the inductor,$L = 40 	ext{ mH} = 40 	imes 10^{-3} 	ext{ H}$.Initial current,$I_1 = 2 	ext{ A}$.Final current,$I_2 = 12 	ext{ A}$.Time interval,$dt = 8 	ext{ ms} = 8 	imes 10^{-3} 	ext{ s}$.The magnitude of the induced emf in an inductor is given by the formula:$|\varepsilon| = L \frac{di}{dt}$Substituting the values:$|\varepsilon| = (40 	imes 10^{-3} 	ext{ H}) 	imes \frac{(12 	ext{ A} - 2 	ext{ A})}{8 	imes 10^{-3} 	ext{ s}}$$|\varepsilon| = 40 	imes 10^{-3} 	imes \frac{10}{8 	imes 10^{-3}}$$|\varepsilon| = 40 	imes \frac{10}{8} = 5 	imes 10 = 50 	ext{ V}$.

The current in an inductor of self-inductance $L=40 \text{ mH}$ is to be increased uniformly from $2 \text{ A}$ to $12 \text{ A}$ in $8 \text{ ms}$. The emf induced in the inductor during this process is (in $\text{ V}$)

Similar Questions

The current $(I)$ in an inductor is varying with time $(t)$ as shown in the figure. Which of the following graphs shows the correct variation of voltage $(V)$ with time $(t)$ in the inductor?

Self-induction of a solenoid cannot be increased by

The self-inductance of a coil is $50 mH$. When a current of $1 A$ passing through the coil reduces to zero at a steady rate in $0.1 s$,find the self-induced emf. (in $V$)

The self-inductance of a solenoid is:

The length of thin wire required to manufacture a solenoid of inductance $L$ and length $\ell$ (if cross-sectional diameter is considered much less than its length) is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module