A coil has an inductance of 2 , H and resista | vedclass

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

Question

(A) The equilibrium current $I$ in the coil is given by Ohm's law: $I = \frac{V}{R} = \frac{10 \, V}{4 \, \Omega} = 2.5 \, A$.The energy $E$ stored in

Vedclass · Accepted Answer

$625$

Vedclass · Answer

(A) The equilibrium current $I$ in the coil is given by Ohm's law: $I = \frac{V}{R} = \frac{10 \, V}{4 \, \Omega} = 2.5 \, A$.The energy $E$ stored in the magnetic field of an inductor is given by the formula: $E = \frac{1}{2} L I^2$.Substituting the given values $L = 2 \, H$ and $I = 2.5 \, A$:$E = \frac{1}{2} 	imes 2 	imes (2.5)^2 = 6.25 \, J$.To express this in the form $.......... 	imes 10^{-2} \, J$:$E = 625 	imes 10^{-2} \, J$.Thus,the value is $625$.

$A$ coil has an inductance of $2 \, H$ and resistance of $4 \, \Omega$. $A$ $10 \, V$ potential difference is applied across the coil. The energy stored in the magnetic field after the current has built up to its equilibrium value will be $.......... \times 10^{-2} \, J$.

Similar Questions

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

The magnetic energy stored in an inductor of inductance $5 \mu H$ carrying a current of $2 \ A$ is

If the current in a coil is halved,then the energy stored becomes how many times the previous value?

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

$A$ $100\, mH$ coil carries a current of $1\, A$. The energy stored in its magnetic field is......$J$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module