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An:- Given data?\rightarrow initially there is no enogry stored in the ciscuit.\rightarrow For\delta(t) \longrightarrow \begin{aligned}t<0 & \Rightarrow 0 \\t>0 & =0 \\t & =0 \Rightarrow 1\end{aligned}\therefore a) Find i_{1} for t \geq 0^{+}:-At t \geq 0^{t} \Rightarrow \delta(t)=1\rightarrow The above diagram can be redrawn asfor Transforma connections:-Notewfor \omega=1 \mathrm{rad} (sec'-\rightarrow Apply KVL in Loop -( :-20+\frac{1}{2} i_{1}+\frac{1}{2}\left(i_{1}-i_{2}\right)=0 \text { (1) }\rightarrow Apply KVL in loop -(2) :-\frac{1}{2} i_{2}+3 i_{2}-\frac{1}{2}\left(i_{1}-i_{2}\right)=0 \text { (2) }\rightarrow By solving equation (1) & (2) we get 'i, e iz :-\begin{array}{l}\Rightarrow \quad i_{1}-\frac{1}{2} i_{2}=20 \\\Rightarrow \quad-\frac{1}{2} i+4 i_{2}=0\end{array}multiply equation (1) by \frac{1}{2} and add resultant to equation (2) we gat?\begin{array}{c}\frac{1}{2} i_{1}-\frac{1}{4} i_{2}=10-\text { (1) } \\\frac{1}{2} i_{1}+4+i_{2}=0 \\4 \frac{1}{2}-\frac{1}{4} i_{2}=10 \\i_{2}\left(\frac{\frac{15}{4}}{4}\right)=10^{2} \\i_{2}=8 / 3 \text { Amps }\end{array}\rightarrow substitute (3) in (1) we get:-\begin{aligned}i_{1} & =20+\frac{1}{2} i 2 \\& =20+\frac{1}{2} \times(8 / 3) \\& =20+8 / 6 \\R_{1} & =\frac{128}{6} \text { Arrps }\end{aligned}\therefore a) R_{1} for t \geq 0^{+} \Rightarrow R_{1}=\frac{128}{6} Anys{ }_{q_{1}}=21.33 \text { Aorys }()\therefore b) i_{2} for t z 0^{+} \Rightarrow i_{2}=8 /_{3}=2.66 Aroys\therefore C) Vo for t \geq 0^{+} \RightarrowV_{0}=3 \times 12=3 \times 8 / 3=8 \text { volts } 11 ...