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

$A$ current of $2\,A$ is increasing at a rate of $4\,A/s$ through a coil of inductance $2\,H$. The energy stored in the inductor per unit time is ... $J/s$.

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

$A$ coil with resistance $20\, \Omega$ and inductance $5\, H$ is connected to a $100\, V$ battery. The energy stored in the coil will be .... $J$.

In what form is the energy stored in an inductor?

If a current of $4 \ A$ produces a magnetic flux of $3 \times 10^{-3} \ Wb$ through a coil of $400$ turns,the energy stored in the coil will be: (in $J$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module