Using nodal analysis calculate the current flowing through all the branches in the network shown in Fig. 2.160.
Applying KCL, we can write
8=I_1+I_2or, 8=\frac{V_1}{3}+\frac{V_1-V_2}{2}
or, 5 \mathrm{~V}_1-3 \mathrm{~V}_2=48 (i)
again I_2=I_3+6
or, \frac{V_1-V_2}{2}=\frac{V_2}{4}+6
or, 2 \mathrm{~V}_1-3 \mathrm{~V}_2=24 (ii)
from (ii)
\begin{aligned} 3 \mathrm{~V}_2 & =2 \mathrm{~V}_1-24 \\ \mathrm{~V}_2 & =\frac{2 \mathrm{~V}_1-24}{3} \end{aligned}Substituting V_2 in (i)
5 \mathrm{~V}_1-3\left(\frac{2 \mathrm{~V}_1-24}{3}\right)=48or, \mathrm{V}_1=8 \mathrm{~V}
Putting value of V_1 in (i)
5 \mathrm{~V}_1-3 \mathrm{~V}_2=48or, 5 \times 8-3 V_2= 48
\begin{aligned} & 3 \mathrm{~V}_2=40-48=-8 \\ & \mathrm{~V}_2=-\frac{8}{3} \mathrm{~V} \\ & \mathrm{I}_2=\frac{\mathrm{V}_1-\mathrm{V}_2}{2}=\frac{8-(-8 / 3)}{2}=5.33 \mathrm{~A} \\ & \mathrm{I}_1=\frac{\mathrm{V}_1}{3}=\frac{8}{3}=2.66 \mathrm{~A} \\ & \mathrm{I}_3=\frac{\mathrm{V}_2}{4}=-\frac{8}{3 \times 4}=-\frac{2}{3}=-0.67 \mathrm{~A} \end{aligned}To cross-check
\mathrm{I}_2=\mathrm{I}_3+6=-0.67+6=5.33 \mathrm{~A}