If a current flowing in a coil is reduced to | vedclass

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

Question

(A) The energy $E$ stored in a coil with self-inductance $L$ carrying current $I$ is given by the formula: $E = \frac{1}{2} LI^2$.Let the initial curr

Vedclass · Accepted Answer

$E_2 = \frac{E_1}{4}$

Vedclass · Answer

(A) The energy $E$ stored in a coil with self-inductance $L$ carrying current $I$ is given by the formula: $E = \frac{1}{2} LI^2$.Let the initial current be $I_1$ and the initial energy be $E_1 = \frac{1}{2} LI_1^2$.Let the new current be $I_2 = \frac{I_1}{2}$ and the new energy be $E_2 = \frac{1}{2} LI_2^2$.Taking the ratio of the two energies:$\frac{E_2}{E_1} = \frac{\frac{1}{2} LI_2^2}{\frac{1}{2} LI_1^2} = \left( \frac{I_2}{I_1} \right)^2$.Substituting $I_2 = \frac{I_1}{2}$:$\frac{E_2}{E_1} = \left( \frac{I_1 / 2}{I_1} \right)^2 = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.Therefore,$E_2 = \frac{E_1}{4}$.

If a current flowing in a coil is reduced to half of its initial value,the relation between the new energy $(E_2)$ and the original energy $(E_1)$ stored in the coil will be

Similar Questions

If the current in an $LR$ circuit is reduced to half,what will be the energy stored in it?

The energy stored in a $50 \, mH$ inductor carrying a current of $4 \, A$ will be $J$.

$A$ coil of self-inductance $0.5 \ mH$ carries a current of $2 \ A$. The energy stored in Joule is . . . . . . .

$A$ coil of inductance $2\, H$ having negligible resistance is connected to a source of supply whose voltage is given by $V = 3t$ volt (where $t$ is in seconds). If the voltage is applied when $t = 0$,then the energy stored in the coil after $4\, s$ is ............ $J$.

In an inductor of inductance $L = 100 \, mH$,a current of $I = 10 \, A$ is flowing. The energy stored in the inductor is ....... $J$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module