Chapter 21, Problem 50P

To determine

The magnitude of $\Delta I/\Delta T$ at time $t=0$, and also show that the current reaches its maximum value in one time constant

Answer to Problem 50P

Solution:

The magnitude of $\Delta I/\Delta T$ at time $t=0$ is $0$

It has been showed that the current will reach its maximum value in one time constant.

Explanation of Solution

Given:

LR circuit diagram

Figure.1

Formula used:

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

$I=I_o\left(1-e^{-t/\tau }\right)$

Here,

$I$ is the current at any time $t$

$I_o$ is the initial current of the circuit

$t$ is the time

$\tau$ is the time constant

Calculation:

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

$I=I_o\left(1-e^{-t/\tau }\right)$

Differentiate the above equation with respect to time

$\begin{array}{c}\frac{dI}{dt}=I_o\left(0-\left(\frac{t}{\tau }\right)e^{-t/t}\right)\\ =I_o\left(\frac{t}{\tau }\right)e^{-t/t}\end{array}$ …… (1)

Now substitute $t=0$ in the above equation

$\begin{array}{c}\frac{dI}{dt}=I_o\left(\frac{0}{\tau }\right)e^{-0/t}\\ =0\end{array}$

To determine the maximum value of Current

$\frac{dI}{dt}=0$

Substitute the values from Eq. (1)

$\begin{array}{c}{I}_o\left(\frac{t}{\tau }\right)e^{-t/t}=0\\ e^{-t/\tau }=0\\ e^{-t/\tau }=\ln 1\\ -\frac{t}{\tau }=e^{\ln 1}\\ -\frac{t}{\tau }=1\\ t=-\tau \\ t=\tau \left(\because \text{time}\text{can}\text{not}\text{be}\text{negative}\right)\end{array}$

Therefore if the current will keep increasing at this rate the current will reach it’s maximum value in one time constant.