Question 2.50: Using nodal analysis calculate the current flowing through a......

Using nodal analysis calculate the current flowing through all the branches in the network shown in Fig. 2.160.

figure 2.160
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Applying KCL, we can write

8=I_1+I_2

or,                   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)=48

or,                                  \mathrm{V}_1=8 \mathrm{~V}

Putting value of V_1 in (i)

5 \mathrm{~V}_1-3 \mathrm{~V}_2=48

or,                      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}

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