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$\text { An unchanged } 0.2 \mu \mathrm{F} \text { capacitor is driven by a triangular currentpulse. The current pulse is described by }$

$\begin{array}{l}{\text { Derive the expressions for the capacitor voltage, power, }} \\ {\text { and energy for each of the four time intervals }} \\ {\text { needed to describe the current. }}\end{array}$

$i(t)=\left\{\begin{array}{ll}{0,} & {t \leq 0} \\ {5000 t A,} & {0 \leq t \leq 20 \mu \mathrm{s}} \\ {0.2-5000 t A,} & {20 \leq t \leq 40 \mu \mathrm{s}} \\ {0,} & {t \geq 40 \mu \mathrm{s}}\end{array}\right.$

$V(t)=\frac{1}{c} \int_{t_{0}}^{t} i(t)dt+V(t_{0})$

$t=0 \rightarrow V{(0)}=0*t_0=0$

$V(t)=\frac{1}{0.2+10^{-6}} \int^{t}_{0} i(t) d t+V(0)$

$V(t)=\frac{1}{0.2+10^{-6}} \int^{t}_{0} i(t) d t+0$

$t \leq 0$                $i (t)=0$

$V(t) \frac{-1}{c}\int_{6}^{t}0=0$

$P(t)=V(t) \ i(t)=0$

$W(t)=\frac{1}{2} C V^2=\frac{1}{2}C*0=0$

$0 \leq t \leq 20 \mu \mathrm{s}$            $i(T)=5000 t$

$V(t)=\frac{1}{0.2 * 10^{-6}}\int_{0}^{t}5000t \ dt$

$V(t)=12.5 * 10^{9} t^2 \longrightarrow (V)$

$P(t)=i(t) \ v(t)=5000t*12.5*10^9 t^2$

$P(t)=62.5*10^{12}t^3 \longrightarrow (W)$

$W(t)=\frac{1}{2} C V^{2}=\frac{1}{2}*0.2*10^{-6} [12.5*10^9t^2]^2$

$W(t)=15.625 * 10^{12} t^{4}$        (Joule)

$20\mu \mathrm{s} \leq t \leq 40 \mu \mathrm{s}$

$V(20 \mu \mathrm{s})=12.5*10^9[20*10^{-6}]^2$

$V(20 \mu \mathrm{s})=5 volt \quad V(t_0)$

$V(t)=\frac{1}{C} \int^t_{20 \mu s} 0.2-5000 t+V(20 \mu s)$

$V(t)=\frac{1}{0.2*10^{-6}} \int^{t}_{20*10^{-6}}(0.2-5000t)dt+5$

$V(t)=10^{6}+-12.5 * 10^{9} t^{2}-10$

$P(t)=V(t) i(t)=62.5*10^{12}t^3-7.5*10^9t^2+2.5*10^5t-2$

$W(t)=\frac{1}{2} CV^{2}=\frac{1}{2}*0.2*10^{-6}*[\cdots]^2$

$W(t)=15.625*10^{12} t^{4}-2.5*10^9t^3+0.125*10^6t^2-2t*10^{-5}$            (Joule)

$t \geq 40 \mu \mathrm{s}$                                    $i(t)=0$

$V(t)=\int^{t}_{40 \mu s} 0dt+ V(40 \mu s)$

$V(40 \mu s)=10 V$

$V(t)=10 V$

$P(t)=0$

$W(t)=\frac{1}{2} C V^{2}=\frac{1}{2} * 0.2*10^{-6}*(10)^2$

$W(t)=10 \mu J$