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Question Number 148825 by DELETED last updated on 31/Jul/21

Answered by DELETED last updated on 31/Jul/21

$$\mathrm{\Sigma }\mathrm{IR}+\mathrm{\Sigma }\mathrm{E}=0$$$${\mathrm{i}}_1+{\mathrm{i}}_2={\mathrm{i}}_3\to {\mathrm{i}}_2={\mathrm{i}}_3-{\mathrm{i}}_2......\left(1\right)$$$${2\mathrm{i}}_1+{3\mathrm{i}}_3+9-6=0$$$${2\mathrm{i}}_1+{3\mathrm{i}}_3=-3....\left(2\right)$$$$$$$${6\mathrm{i}}_2+{3\mathrm{i}}_3+9-12=0$$$${6\mathrm{i}}_2+{3\mathrm{i}}_3=3$$$$6({\mathrm{i}}_3-{\mathrm{i}}_1)+{3\mathrm{i}}_3=3$$$$-{6\mathrm{i}}_1+{9\mathrm{i}}_3=3....\left(3\right)$$$$\left(2\right)\to \left(3\right){2\mathrm{i}}_1+{3\mathrm{i}}_3=-3\times 3$$$$-{6\mathrm{i}}_1+{9\mathrm{i}}_3=3$$$$-------+$$$${6\mathrm{i}}_1+{9\mathrm{i}}_3=-9$$$$-{6\mathrm{i}}_1+{9\mathrm{i}}_3=3$$$$-------+$$$${18\mathrm{i}}_3=-6$$$${\mathrm{i}}_3=-\frac{1}{3}\mathrm{A}$$$${2\mathrm{i}}_1+{3\mathrm{i}}_3=-3$$$${2\mathrm{i}}_1-1=-3$$$${2\mathrm{i}}_1=-2\to {\mathrm{i}}_1=-1\mathrm{A}$$$${\mathrm{i}}_2=-\frac{1}{3}+1=\frac{2}{3}\mathrm{A}$$$$$$$$$$$$$$$$$$

Answered by JDamian last updated on 31/Jul/21

$$\begin{array}{c}{V}_{AB}+V_{BC}=V_{AC}\\ V_{DB}+V_{BC}=V_{DC}\end{array}\}$$$$$$$$\begin{array}{c}2I_x+3(I_x+I_y)+9=6\\ 6I_y+3(I_x+I_y)+9=12\end{array}\}\Rightarrow \begin{array}{c}5I_x+3I_y=-3\\ 3I_x+9I_y=3\end{array}\}$$$$$$$$I_x=\frac{\left|\begin{array}{cc}-3& 3\\ 3& 9\end{array}\right|}{\left|\begin{array}{cc}5& 3\\ 3& 9\end{array}\right|}=\frac{-27-9}{45-9}=\frac{-36}{36}=-1\mathrm{A}$$$$I_y=\frac{\left|\begin{array}{cc}5& -3\\ 3& 3\end{array}\right|}{\left|\begin{array}{cc}5& 3\\ 3& 9\end{array}\right|}=\frac{15+9}{45-9}=\frac{24}{36}=\frac{2}{3}\mathrm{A}$$

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