Question 2.47: For the circuit shown in Fig. 2.155, calculate the current f......
For the circuit shown in Fig. 2.155, calculate the current flowing through the 5 Ω resistor by using the nodal method.
Let V_A\ and\ V_B be the potentials at node A and node B, respectively. Let the reference node be at C. Let us assume current directions at node A as shown in Fig. 2.156.
We will have incoming currents as equal to outgoing current, i.e.,
I_1+I_2=I_3 (i)
Current, \mathrm{I}_1=\left(\mathrm{V}_{\mathrm{P}}-\mathrm{V}_{\mathrm{A}}\right) / \mathrm{R}=\frac{6-\mathrm{V}_{\mathrm{A}}}{2} (ii)
Current, \mathrm{I}_2=\frac{5-\mathrm{V}_{\mathrm{A}}}{\mathrm{R}}=\frac{5-\mathrm{V}_{\mathrm{A}}}{2} \quad[\because \text { potential of point } \mathrm{T} \text { is }+5 \mathrm{~V}] (iii)
Current, \mathrm{I}_3=\frac{\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}}{\mathrm{R}}=\frac{\mathrm{V}_{\mathrm{A}}-(-8)}{5} (iv)
Note: Potential of point B with respect to C is –8 V.
Therefore, from (i), (ii), (iii) and (iv),
\frac{6-\mathrm{V}_{\mathrm{A}}}{2}+\frac{5-\mathrm{V}_{\mathrm{A}}}{2}-\frac{\mathrm{V}_{\mathrm{A}}+8}{5}=0or, \frac{5\left(6-V_A\right)+5\left(5-V_A\right)-2\left(V_A+8\right)}{10}=0
or, -12 \mathrm{~V}_{\mathrm{A}}+39=0
or, \mathrm{V}_{\mathrm{A}}=\frac{39}{12}=3.25 \mathrm{~V}
Current through the 5 Ω resistor is I_3.
\begin{aligned} \mathrm{I}_3 & =\frac{\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}}{5}=\frac{3.25-(-8)}{5} \\ & =\frac{3.25+8}{5}=2.25 \mathrm{~A} \end{aligned}