Question

For $L-R$ circuit, the time constant is equal to-

• A. Twice the ratio of the energy stored in the magnetic field to the rate of the dissipation of energy in the resistance.
• B. The ratio of the energy stored in the magnetic field to the rate of dissipation of energy in the resistance.
• C. Half of the ratio of the energy stored in the magnetic field to the rate of dissipation of energy in the resistance.
• D. The square of the ratio of the energy stored in the magnetic field to the rate of dissipation of energy in the resistance.

Answer 1

The correct option is A Twice the ratio of the energy stored in the magnetic field to the rate of the dissipation of energy in the resistance.

We know that the magnetic field energy stored in an $L-R$ circuit is,

$U_B=\frac{1}{2}L{i}^2.....\left(1\right)$

Where,
$L=$ inductance of the circuit and,
$i=$ current flowing through the circuit

Rate of dissipation of energy across the resistor $R$ is, $P=\frac{dU}{dt}$

$P=i^{2}R.......\left(2\right)$

The time constant of the circuit is,

$\tau =\frac{L}{R}..........\left(3\right)$

From $\left(1\right),\left(2\right)$ and $\left(3\right)$ we get,

$\frac{U_B}{P}=\frac{\frac{1}{2}L{i}^2}{i^{2}R}=\frac{1}{2}\frac{L}{R}=\frac{\tau }{2}$

$\therefore \tau =\frac{2U_B}{P}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(\text{A}\right)$ is the correct answer.