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Given circuitand P=110 \sin 2 t=110 \cos \left(2 t-90^{\circ}\right) \text {; Here } \omega=2 \mathrm{rad} / \mathrm{sec}in phasor P=110 \angle-90^{\circ}Convert the given circuit into phasor.Let i_{1} and is be flowing through Two loops.Thisccise Mutual Inductance is aiding. convert the given Transformer into T-Nefwork.\begin{array}{l}\text { CF } \equiv-1 / j \omega c n \\\end{array}This circuit cambe written asZeq.\begin{aligned}\text { Here zea } & =(j 3-j 1) /(j \\& =j 21 l j \\& =\frac{j 2 \times j}{j 2+j}=\frac{2}{3} j\end{aligned}voltage Across zea is Applyvoltage division rule\begin{aligned}V_{\text {zea }} & =\frac{110<-90 \times 2 / 3 j}{2 / 3 j+j 3} \\& =110<-90^{\circ} \times \frac{2 / 7}{11 / 3} \\V_{\text {zeq }} & =20<-90^{\circ} \mathrm{V}\end{aligned}Apply voltage division rule\begin{array}{l}\text { Then } V_{0}=V_{z e 9} \frac{-j 1}{-j 1+j 3}=20 \angle-90^{\circ} \times \frac{-j 1}{j \nu} \\=-10<-90^{\circ} \\V_{0}=-10 \angle-90^{\circ} \text { votts } \\\end{array}Then v_{0}(t)=-10 \cos (2 t-90)\begin{array}{l}=-10 \sin 2 t \\=10 \sin \left(2 t+180^{\circ}\right)\end{array}But answer is in cosine form\begin{aligned}\therefore V_{0}(t) & =10 \cos \left(2 t+180^{\circ}-90^{\circ}\right) \\v_{0}(t) & =10 \cos \left(2 t+90^{\circ}\right) v_{0} 1 t s\end{aligned} ...