Chapter 7.6, Problem 20P

Repeat Problem 19, except suppose that the switch is alternately closed and opened at times $t=0,\frac{\pi }{5},\frac{2\pi }{5},\mathrm{....}\frac{n\pi }{5}$

Now show that if

$\frac{n\pi }{5}<t<\frac{\left(n+1\right)\pi }{5}$, then $i\left(t\right)=\{\begin{array}{cc}\sin 10t& \text{if }n\text{ is even;}\\ 0& \text{if }n\text{ is odd}\text{.}\end{array}$

Thus the current in alternate cycles of length $\frac{\pi }{5}$ first exe cutes a sine oscillation during one cycle, then is dormant during the next cycle, and so on (see Fig. 7.6.8).