Problem

# Find Vo and V1 in Fig. E2.23.Figure E2.23

Find Vo and V1 in Fig. E2.23.

Figure E2.23

#### Step-by-Step Solution

Solution 1

Refer to Figure E2.23 in the textbook.

Represent the currents on the circuit.

Apply Kirchhoff current law at node $$a$$.

$$I_{1}=I_{2}+I_{5}+I_{6}$$

$$\frac{0-V_{a}}{4 \mathrm{k} \Omega}=\frac{V_{a}-V_{b}}{10 \mathrm{k} \Omega}+\frac{V_{a}-0}{6 \mathrm{k} \Omega}+\frac{V_{a}-0}{12 \mathrm{k} \Omega}$$

$$\frac{V_{a}}{4 \mathrm{k} \Omega}+\frac{V_{a}-V_{b}}{10 \mathrm{k} \Omega}+\frac{V_{a}}{6 \mathrm{k} \Omega}+\frac{V_{a}}{12 \mathrm{k} \Omega}=0$$

$$\frac{V_{a}}{4 \mathrm{k} \Omega}+\frac{V_{a}}{10 \mathrm{k} \Omega}+\frac{V_{a}}{6 \mathrm{k} \Omega}+\frac{V_{a}}{12 \mathrm{k} \Omega}=\frac{V_{b}}{10 \mathrm{k} \Omega}$$

$$V_{a}\left(\frac{1}{4 \mathrm{k} \Omega}+\frac{1}{10 \mathrm{k} \Omega}+\frac{1}{6 \mathrm{k} \Omega}+\frac{1}{12 \mathrm{k} \Omega}\right)=\frac{V_{b}}{10 \mathrm{k} \Omega}$$

$$V_{a}\left(\frac{15+6+10+5}{60 \mathrm{k} \Omega}\right)=\frac{V_{b}}{10 \mathrm{k} \Omega}$$

$$V_{a}(15+6+10+5)=\left(\frac{60 \mathrm{k} \Omega}{10 \mathrm{k} \Omega}\right) V_{b}$$

$$36 V_{a}=6 V_{b}$$

$$V_{b}=6 V_{a}$$

Apply Kirchhoff current law at node b.

$$I_{2}=I_{3}+I_{4}$$

$$\frac{V_{a}-V_{b}}{10 \mathrm{k} \Omega}=20 \mathrm{~mA}+\frac{V_{b}-0}{4 \mathrm{k} \Omega} \quad\left(\right.$$ since, $$\left.I_{3}=20 \mathrm{~mA}\right)$$

$$\frac{V_{a}-V_{b}}{10 \mathrm{k} \Omega}-\frac{V_{b}}{4 \mathrm{k} \Omega}=20 \mathrm{~mA}$$

$$\frac{2 V_{a}-2 V_{b}-5 V_{b}}{20 \mathrm{k} \Omega}=20 \mathrm{~mA}$$

$$2 V_{a}-2 V_{b}-5 V_{b}=(20 \mathrm{~mA})(20 \mathrm{k} \Omega)$$

$$2 V_{a}-7 V_{b}=400$$

Substitute the expression of $$V_{b}$$.

$$2 V_{a}-7 V_{b}=400$$

$$2 V_{a}-7\left(6 V_{a}\right)=400$$

$$2 V_{a}-42 V_{a}=400$$

$$-40 V_{a}=400$$

$$V_{a}=\frac{400}{-40}$$

$$=-10 \mathrm{~V}$$

Substitute the value of $$V_{a}$$ in the equation of $$V_{b}$$.

$$V_{b}=6 V_{a}$$

$$=6(-10 \mathrm{~V})$$

$$=-60 \mathrm{~V}$$

Find the voltages represented on the circuit.

\begin{aligned} V_{o} &=V_{b} \\ &=-60 \mathrm{~V} \\ V_{1} &=0-V_{a} \\ &=0-(-10 \mathrm{~V}) \\ &=10 \mathrm{~V} \end{aligned}

Thus, the voltages $$V_{o}$$ and $$V_{1}$$ are $$-60 \mathrm{~V}$$ and $$10 \mathrm{~V}$$ respectively.