Question

An inductor coil stores $32\text{J}$ of magnetic field energy and dissipates energy as heat at the rate of $320\text{W}$ when the instant current is $4\text{A}$ through it. Find the time constant of the circuit.

• A. $2\text{s}$
• B. $0.2\text{s}$
• C. $0.02\text{s}$
• D. $0.002\text{s}$

Answer 1

The correct option is B $0.2\text{s}$

The magnetic field energy stored in the inductor at the given instant is,

$U=\frac{1}{2}L{i}^2$

$\Rightarrow 32=\frac{1}{2}L(4{)}^2$

$\Rightarrow L=4\text{H}$

The power dissipated as heat at the given instant is,

$P=i^{2}R$

$\Rightarrow 320=4^{2}R$

$\Rightarrow R=20\Omega$

The time constant of the circuit is,

$\tau =\frac{L}{R}=\frac{4}{20}=0.2\text{s}$

Hence, option $\left(B\right)$ is the correct answer.