Question

A $50{\text{ mH}}$ coil carries a current of $2$ amp, the energy stored in joule is:
(A) $1$
(B) $0.05$
(C) $0.1$
(D) $0.5$

Answer 1

Recall the concept of energy stored in an inductor and the formula for calculating it and put the values of inductance and current as given in question.

Formula used:
$E = \dfrac{1}{2}L{I^2}$
Here $L$ is the Inductance of the coil and $I$ be the current flowing through it.

Complete step by step answer:
The inductance of the coil, $L = 50mH$
Convert the given unit in the MKS (Metre Kilogram Second) system of units.
$1mm = {10^{ - 3}}m$
Place the value in the inductance
$L = 50 \times {10^{ - 3}}H$
Also, give that the current passing through the coil, $I = 2A$
Energy stored in an inductor, $E = \dfrac{1}{2}L{I^2}$
Place the known values in the above equation –
$E = \dfrac{1}{2} \times 50 \times {10^{ - 3}} \times 2 \times 2$
Simplify the above equation using the basic mathematical operations. Remove $2$ from the denominator and the numerator as they both cancel each other.
$E = 50 \times {10^{ - 3}} \times 2$
Simplify
$E = 100 \times {10^{ - 3}} \\\ \therefore E = 0.1J$
Therefore, the energy stored in joule is $0.1J$ .Hence,the option C is the correct answer.

Note: Inductance can be defined as the flux linkage with the line per unit carrying current flowing through it. The units of the inductors are measured in Henry, milli-Henry and Micro Henry. Also, refer to the concepts of resistance and capacitance and know the difference between RLC (Resistance, Inductor and Capacitor) properly.