Chapter 4.4, Problem 40E

To determine

To find: The maximum or peak current.

Answer to Problem 40E

The maximum or peak current is $i=2\sqrt{2}\text{amp}$ .

Explanation of Solution

Given:

The current in a circuit is given by $i=2\cos t+2\sin t$ .

Calculation:

Differentiate the given function with respect to t and equate it with 0 and solve for t .

  $\begin{array}{l}\frac{di}{dt}=\frac{d}{dt}(2\cos t+2\sin t)\\ \frac{di}{dt}=-2\sin t+2\cos t\end{array}$

Substituting $\frac{di}{dt}=0$ in the above equation.

  $\begin{array}{c}0=-2\sin t+2\cos t\\ 2\sin t=2\cos t\\ \sin t=\cos t\end{array}$

As

  $\begin{array}{c}{\sin }^{2}t+{\cos }^{2}t=1\\ {\sin }^{2}t={\cos }^{2}t=\frac{1}{2}\\ \sin t=\cos t=\frac{1}{\sqrt{2}}\end{array}$

It occurs at $t=\frac{k\pi }{4}$ , where k is any odd integer.

As,

  $\begin{array}{l}\sin t=\cos t\\ t=\frac{\pi }{4}+m\pi \end{array}$

where m is any integer.

Differentiate $\frac{di}{dt}$ with respect to t .

  $\begin{array}{c}\frac{d^{2}i}{d{t}^2}=\left(-2\sin t+2\cos t\right)\text{'}\\ =-2\cos t-2\sin t\end{array}$

Using $\sin t=\cos t=\frac{1}{\sqrt{2}}$

  $\begin{array}{c}\frac{d^{2}i}{d{t}^2}=-2\left(\frac{1}{\sqrt{2}}\right)-2\left(\frac{1}{\sqrt{2}}\right)\\ =-4\left(\frac{1}{\sqrt{2}}\right)<0\end{array}$

By second derivative test, the maximum value of the function occurs at $t=\frac{\pi }{4}+m\pi$ , where m is any integer.

Now,

  $\begin{array}{l}i=2\sin t+2\cos t\\ i=2\frac{1}{\sqrt{2}}+2\frac{1}{\sqrt{2}}\\ i=2\sqrt{2}\end{array}$

The maximum or peak current is $i=2\sqrt{2}\text{ amp}$ .