The current in an inductor is given by I = (3 | vedclass

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

Question

(B) Given,the current in the inductor is $I = (3t + 8) \ A$.The induced emf $\varepsilon$ in an inductor is given by the formula $\varepsilon = -L \fr

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Given,the current in the inductor is $I = (3t + 8) \ A$.The induced emf $\varepsilon$ in an inductor is given by the formula $\varepsilon = -L \frac{dI}{dt}$.The magnitude of the induced emf is $|\varepsilon| = L \left| \frac{dI}{dt} ight|$.Given $|\varepsilon| = 12 \ mV = 12 	imes 10^{-3} \ V$.First,calculate the rate of change of current: $\frac{dI}{dt} = \frac{d}{dt}(3t + 8) = 3 \ A/s$.Now,substitute the values into the formula: $12 \ mV = L 	imes 3 \ A/s$.$L = \frac{12 \ mV}{3 \ A/s} = 4 \ mH$.Therefore,the self-inductance of the inductor is $4 \ mH$.

The current in an inductor is given by $I = (3t + 8)$ $A$,where $t$ is in seconds. The magnitude of the induced emf produced in the inductor is $12 \ mV$. The self-inductance of the inductor is . . . . . . $mH$.

Similar Questions

The adjoining figure shows two bulbs $B_1$ and $B_2$,a resistor $R$,and an inductor $L$. When the switch $S$ is turned off,what happens to the bulbs?

What is the self-inductance of a solenoid of length $31.4 \ cm$,area of cross-section $10^{-3} \ m^2$,and total number of turns $10^3$?

The self-inductance of a solenoid of length $L$,area of cross-section $A$,and having $N$ turns is:

The self-inductance of a coil is $5 \, H$. If a current of $1 \, A$ changes to $2 \, A$ within $5 \, s$ through the coil,the magnitude of the induced $e.m.f.$ will be $... \, V$.

If a coil is open,then its self-inductance $L$ and resistance $R$ become:

Vedclass Test Series

Exam Paper Generator

Online Exam Module