Question

When a battery, a resistor, a switch, and an inductor form a circuit and the switch is closed, the inductor acts to oppose the change in the current.
How is the time constant of the circuit affected by doubling the inductance in the circuit?

Answer 1

Answer:

The time constant becomes twice.

Explanation:

$T$ = Time constant of the L-R circuit

$L$ = Inductance of the inductor

$R$ = Resistance of the resistor

Time constant of the L-R circuit is given as

$T = \frac{L}{R}$

$T_{1}$ = initial time constant of the L-R circuit = $T$

$T_{2}$ = final time constant of the L-R circuit

$L_{1}$ = Initial inductance of the inductor =  $L$

$L_{2}$ = Initial inductance of the inductor =  $2L$

For the same resistance, the time constant depend directly on the inductance, hence

$\frac{T_{1}}{T_{2}} = \frac{L_{1}}{L_{2}}\\\frac{T}{T_{2}} = \frac{L}{2L}\\\frac{T}{T_{2}} = \frac{1}{2}\\T_{2} = 2T$