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Solution 3) Griven c=200 \mu \mathrm{F}(a)\begin{array}{c}v(t)=0 ; \begin{array}{cc}t<0 \\t \\t>8\end{array} \\v(t)=\left\{\begin{array}{cc}0 ; t)=10 t \\10 t ; & 0 \leq t \leq 2 \\-10 t+40 ; & 2 \leq t \leq 6 \\10 t-80 ; & 6 \leq t \leq 8\end{array}\right.\end{array}v(t)=0 ;+<0 \quad|v(t)=10 t ; 0 \leq t \leq 2| v(t)=-10 t+40 ; 2 \leq t \leq 6 \mid v(t)=10 t-80 ; \quad 6 \leq t \leq 8(b)\begin{array}{l}\text { cosrent }\left(I_{c}\right)=c \frac{d V(t)}{d t}=\left(0.2 \times 10^{-3}\right) \frac{d V(t)}{d t}=0.2 \frac{d V(t)}{d t} \mathrm{~mA} \\I_{c}(t)=\left\{\begin{array}{ll}0 \mathrm{~mA} ; & t<0 \quad \& t>8 \\2 \mathrm{~mA} ; & 0 \leq t \leq 2 \\-2 \mathrm{~mA} ; & 2 \leq t \leq 6 \\2 \mathrm{~mA} ; & 6 \leq t \leq 8\end{array}\right.\end{array}\begin{array}{l}\text { Power } P(t)=V(t) \cdot I_{c}(t)=\left\{\begin{array}{l}0 m \omega ; t<0 \& t>8 \\20 t m \omega ; 0 \leq t \leq 2 \\20 t-80 \mathrm{mw} j 2 \leq t \leq 6 \\20 t-160 \mathrm{mw} ; \sigma \leq t \leq 8\end{array}\right. \\{\left[\text { Energy } E(t)=\int p(t) d t\right] \quad \text { Hence } E(t)=\left\{\begin{array}{c}0 m j ; t<0 \\40 \mathrm{~mJ} ; 0 \leq t \leq 2 \\0 ; 2 \leq t \leq 6 \\-300 \mathrm{~mJ} ; t \geqslant 6\end{array}\right.} \\\end{array}(c) Power is absorbed when it is positive and when it is delivered it is negatine. since in no time enterval the power is not negative for 0 to 2 it is negative for 2<t<4 \mathrm{sec} and it is negatine for 6 \leq t \leq 8.Hence power is delivered for 2<t<4 and 6<t<b.Power is absorbed for 0 \leq t \leq 2 and 4 \leq t \leq 6 ...