Chapter 21, Problem 49P

To determine

Time constant takes for the potential difference across the resistor in an LR circuit to drop 2.5% of its original value

Answer to Problem 49P

Solution:

Time constant required is $3.7$.

Explanation of Solution

Given:

Drop in voltage, $V=2.5\%$ of its original value, $V=0.025V_o$

LR circuit diagram

Figure.1

Formula used:

In LR circuit the voltage at any time $t$ is given as

$V=V_o{e}^{-t/\tau }$

Here,

$V$ is the voltage at any time $t$

$V_o$ is the initial voltage of the circuit

$t$ is the time

$\tau$ is the time constant

Calculation:

The time constant is calculated as

$V=V_o{e}^{-t/\tau }$

Plugging the values in the above equation

$\begin{array}{c}0.025V_o=V_o{e}^{-t/\tau }\\ 0.025=e^{-t/\tau }\\ -\frac{t}{\tau }=\ln 0.025\\ -\frac{t}{\tau }=-3.7\\ t=3.7\tau \end{array}$