Use nodal analysis to solve for the nodal voltages in the circuit of Fig. E.4.
At node 1,
2 = \frac{V_{1} – V_{2}}{4} + \frac{V_{1} – 0}{8} → 16 = 3 V_{1} – 2 V_{2} (E.2.1)
At node 2,
3I_{x} = \frac{V_{2} – V_{1}}{4} + \frac{V_{2} – V_{3}}{2} + \frac{V_{2} – V_{4}}{2}
But
I_{x} = \frac{V_{4} – V_{3}}{4}
so that
3( \frac{V_{4} – V_{3}}{4} ) = \frac{V_{2} – V_{1}}{4} + \frac{V_{2} – V_{3}}{2} + \frac{V_{2} – V_{4}}{2} → (E.2.2)
0 = – V_{1} + 5 V_{2} + V_{3} – 5 V_{4}
At node 3,
3 = \frac{ V_{3} – V_{2}}{2} + \frac{V_{3} – V_{4}}{4} → 12 = – 2V_{2} + 3V_{3} – V_{4} (E.2.3)
At node 4,
0 = 2 + \frac{V_{4} – V_{2}}{2} + \frac{V_{4} – V_{3}}{4} → – 8 = – 2 V_{2} – V_{3} + 3 V_{4} (E.2.4)
Combining Eqs. (E.2.1) to (E.2.4) gives
\begin{bmatrix}3 & -2 & 0 & 0 \\ -1 & 5 & 1 & -5 \\ 0 & -2 & 3 & -1 \\ 0 & -2 & -1 & 3 \end{bmatrix} \begin{bmatrix} V_{1} \\ V_{2} \\ V_{3} \\ V_{4} \end{bmatrix} = \begin{bmatrix}16 \\ 0 \\12 \\-8 \end{bmatrix}
or
AV = B
We now use MATLAB to determine the nodal voltages contained in vector .
>> A = [ 3 -2 0 0;
-1 5 1 -5;
0 -2 3 -1;
0 -2 -1 3];
>> B = [16 0 12 -8]’;
>> V = inv(A)*B
V =
-6.0000
-17.0000
-13.5000
-18.5000
Hence V_{1} = - 6.0 , V_{2} = - 17 , V_{3} = - 13.5 , and V_{4} = - 18.5 V .