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 resistance in the circuit?

Answer 1

Answer:Time constant gets doubled

Explanation:

Given

L-R circuit is given and suppose R and L is the resistance and inductance of the circuit then current is given by

$i=i_0\left [ 1-e^{-\frac{t}{\tau }}\right ]$

where $i_0$ is maximum current

i=current at any time

$\tau =\frac{L}{R}=time\ constant$

$\tau '=\frac{2L}{R}=2\tau$

thus if inductance is doubled then time constant also gets doubled or twice to its original value.