Question B.6: Finding the output voltage using the superposition theorem A...

The equivalent circuit with source V_{ S 1} only is shown in Fig. B.9(a). Applying the voltage divider rule, we can find the output voltage due to V_{ S 1} as

V_{ O 1}=\frac{R_{2} \| R_{3}}{R_{1}+R_{2} \| R_{3}} V_{ S 1}=\frac{4 k \| 6 k }{2 k +(4 k \| 6 k )} \times 10=5.45 V

The equivalent circuit with source V_{ S 2} only is shown in Fig. B.9(b); the output voltage due to V_{ S 2} is

V_{ O 2}=\frac{R_{1} \| R_{3}}{R_{2}+R_{1} \| R_{3}} V_{ S 2}=\frac{2 k \| 6 k }{4 k +(2 k \| 6 k )} \times 12=3.27 V

The equivalent circuit with source V_{ S 3} only is shown in Fig. B.9(c); the output voltage due to V_{ S 3} is

V_{ O 3}=\frac{R_{1} \| R_{2}}{R_{3}+R_{1} \| R_{2}} V_{ S 3}=\frac{2 k \| 4 k }{6 k +(2 k \| 4 k )} \times 15=2.73 V

Therefore, the resultant output voltage V_{ O } can be found by combining the output voltages due to individual sources; that is,

V_{ O }=V_{ O 1}+V_{ O 2}+V_{ O 3}=5.45+3.27+2.73=11.45 V

An alternative approach is to apply KVL at node 1 and look for V_{ O }:

\begin{aligned}V_{ O } &=\frac{\text { Currents into node } 1 \text { if it were at ground potential }}{\text { Conductances radiating from node } 1} \\&=\frac{V_{ S 1} / R_{1}+V_{ S 2} / R_{2}+V_{ S 3} / R_{3}}{1 / R_{1}+1 / R_{2}+1 / R_{3}}=\frac{10 / 2 k +12 / 4 k +15 / 6 k }{1 / 2 k +1 / 4 k +1 / 6 k }=11.45 V\end{aligned}