Question 6.5: An uncharged 0.2μF capacitor is driven by a triangular curr...

a) For $t\leq 0$,v,p,and w all are zero.

For $0\leq t\leq 20\mu s$;

$v=5\times 10^{6}=\int_{0}^{t}{\left(5000\tau \right) d\tau } +0=12.5\times10^{9}t^{2 } V$,

$p=vi=62.5\times10^{12}t^{3 } W$,

$w=\frac{1}{2}Cv^{2}= 15.625 \times 10^{12}t^{4} J$.

For $20\mu s\leq t\leq 40\mu s$,

$v=5\times 10^{6}=\int_{20\mu s}^{t}{\left(0.2-5000\tau \right) d\tau } +5$.

(Note that 5 V is the voltage on the capacitor at the end of the preceding interval.) Then,

$v=\left(10^{6}t-12.5\times10^{9}t^{2 }-10\right) V$,

$p=vi$,

$\left(62.5\times10^{12}t^{3 }-7.5\times10^{9}t^{2 }+2.5\times10^{5}t-2\right) W$,

$w=\frac{1}{2}Cv^{2}$,

$= \left(15.625 \times 10^{12}t^{4}-2.5 \times 10^{9}t^{3}+0.125 \times 10^{6}t^{2}-2t+10^{-5}\right) J$.

For $t\geq 40 \mu s$,

$v=10 V$,

$p=vi=0$,

$w=\frac{1}{2}Cv^{2}=10 \mu J$.

b) The excitation current and the resulting voltage, power, and energy are plotted in Fig. 6.12.

c) Note that the power is always positive for the duration of the current pulse, which means that energy is continuously being stored in the capacitor. When the current returns to zero, the stored energy is trapped because the ideal capacitor offers no means for dissipating energy. Thus a voltage remains on the capacitor after i returns to zero.