Chapter 2, Problem 23P

In the circuit shown in Fig. 2.87, determine V_x and the power absorbed by the 60-Ω resistor.

Figure 2.87

For Prob. 2.23.

To determine

Calculate the value of $V_x$ and power absorbed by $60\Omega$ resistor in Figure 2.87.

Answer to Problem 23P

The value of $V_x$ is $\underset{\_}{-100\text{V}}$ and the power absorbed by $60\Omega$ resistor is $\underset{\_}{960\text{W}}$.

Explanation of Solution

Formula used:

Consider the expression for $N$ resistors connected in parallel.

$\frac{1}{R_{\text{eq}}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\cdots +\frac{1}{R_N}$

Here,

$R_1,R_2,R_3\cdots R_N$ are resistors.

Consider the expression for $N$ resistors connected in series.

$R_{\text{eq}}=R_1+R_2+R_3+\cdots +R_N$

Calculation:

Refer to Figure 2.87 in the textbook for prob. 2.23.

In Figure 2.87, $40\Omega \text{and}60\Omega$ are connected in parallel, therefore the equivalent resistance for the parallel connected resistance are calculated as follows.

$\begin{array}{c}\frac{1}{R_{\text{eq}1}}=\frac{1}{40\Omega }+\frac{1}{60\Omega }\\ =\frac{3+2}{120\Omega }\\ =\frac{5}{120\Omega }\\ =\frac{1}{24\Omega }\end{array}$

Simplify the equation as follows.

$\begin{array}{c}{R}_{\text{eq}1}=\frac{1}{\left(\frac{1}{24\Omega }\right)}\\ =24\Omega \end{array}$

As $R_{\text{eq}1}$ and $6\Omega$ are connected in series, therefore the equivalent resistance for the series connected resistance are calculated as follows.

$\begin{array}{c}{R}_{\text{eq}2}=R_{\text{eq}1}+6\Omega \\ =24\Omega +6\Omega \\ =30\Omega \end{array}$

In Figure 2.87, $15\Omega \text{and}30\Omega$ are connected in parallel, therefore the equivalent resistance for the parallel connected resistance are calculated as follows.

$\begin{array}{c}\frac{1}{R_{\text{eq}3}}=\frac{1}{15\Omega }+\frac{1}{30\Omega }\\ =\frac{1+2}{30\Omega }\\ =\frac{3}{30\Omega }\\ =\frac{1}{10\Omega }\end{array}$

Simplify the equation as follows.

$\begin{array}{c}{R}_{\text{eq}3}=\frac{1}{\left(\frac{1}{10\Omega }\right)}\\ =10\Omega \end{array}$

As $R_{\text{eq}3}$ and $20\Omega$ are connected in series, therefore the equivalent resistance for the series connected resistance are calculated as follows.

$\begin{array}{c}{R}_{\text{eq}4}=R_{\text{eq}3}+20\Omega \\ =10\Omega +20\Omega \\ =30\Omega \end{array}$

As $R_{\text{eq}2}\text{and}{R}_{\text{eq}4}$ are connected in parallel, therefore the equivalent resistance for the parallel connected resistance are calculated as follows.

$\begin{array}{c}\frac{1}{R_{\text{eq}5}}=\frac{1}{R_{\text{eq}2}}+\frac{1}{R_{\text{eq}4}}\\ =\frac{1}{30\Omega }+\frac{1}{30\Omega }\\ =\frac{1+1}{30\Omega }\\ =\frac{2}{30\Omega }\end{array}$

Simplify the equation as follows.

$\begin{array}{c}{R}_{\text{eq}5}=\frac{1}{\left(\frac{1}{15\Omega }\right)}\\ =15\Omega \end{array}$

Redraw Figure 2.87 as shown in Figure 1.

Apply KCL at node $V_1$ of Figure 1.

$\begin{array}{l}60\text{A+}\frac{V_1}{10\Omega }+\frac{V_1}{\left(5+R_{\text{eq}5}\right)}=0\\ 60\text{A+}\frac{V_1}{10\Omega }+\frac{V_1}{\left(5+15\Omega \right)}=0\left\{\because R_{\text{eq}5}=15\Omega \right\}\\ 60\text{A+}\frac{V_1}{10\Omega }+\frac{V_1}{20\Omega }=0\\ V_1\left[\frac{1}{10\Omega }+\frac{1}{20\Omega }\right]=-60\text{A}\end{array}$

Simplify the equation as follows.

$\begin{array}{l}{V}_1\left(0.1+0.05\right)\mho =-60\text{A}\\ V_1\left(0.15\mho \right)=-60\text{A}\\ V_1=-\frac{60\text{A}}{0.15\mho }\\ V_1=-400\text{V}\end{array}$

Consider current through $5\Omega$ resistor as $i_5$, from Figure 1 the expression for $i_5$ can be written as follows.

$i_5=\frac{V_1}{5\Omega +R_{\text{eq5}}}$

Substitute $15\Omega$ for $R_{\text{eq5}}$ and $-400\text{V}$ for $V_1$ as follows.

$\begin{array}{c}{i}_5=\frac{-400\text{V}}{5\Omega +15\Omega }\\ =\frac{-400\text{V}}{20\Omega }\\ =-20\text{A}\end{array}$

From Figure 1, consider the expression for $v_x$.

$v_x=\left(5\Omega \right)i_5$

Substitute $-20\text{A}$ for $i_5$ as follows.

$\begin{array}{c}{v}_x=\left(5\Omega \right)\left(-20\text{A}\right)\\ =-100\text{V}\end{array}$

From Figure 1, the voltage across $V_{R_{\text{eq}5}}=V_1-v_x$.

Substitute $-100\text{V}$ for $v_x$ and $-400\text{V}$ for $V_1$ as follows.

$\begin{array}{c}{V}_{R_{\text{eq}5}}=\left(-400\text{V}\right)-\left(-100\text{V}\right)\\ =-400\text{V}+100\text{V}\\ \text{=}-300\text{V}\end{array}$

From voltage division rule, the voltage across $60\Omega$ resistor is calculated as follows.

$V_{60\Omega }=V_{{\mathrm{R}}_{\text{eq5}}}\left(\frac{R_{\text{eq}1}}{6\Omega +R_{\text{eq}1}}\right)$

Substitute $-300\text{V}$ for $V_{R_{\text{eq}5}}$ and $24\Omega$ for $R_{\text{eq}1}$ as follows.

$\begin{array}{c}{V}_{60\Omega }=\left(-300\text{V}\right)\left(\frac{24\Omega }{6\Omega +24\Omega }\right)\\ =\left(-300\text{V}\right)\left(\frac{24\Omega }{30\Omega }\right)\\ =-240\text{V}\end{array}$

Write the expression to find the power absorbed by $60\Omega$ resistor.

$p_{60\Omega }=\frac{{\left(V_{60\Omega }\right)}^2}{60\Omega }$

Substitute $-240\text{V}$ for $V_{60\Omega }$ to obtain the power absorbed by $60\Omega$ resistor.

$\begin{array}{c}{p}_{60\Omega }=\frac{{\left(-240\text{V}\right)}^2}{60\Omega }\\ =\frac{57,600}{60}\text{W}\\ =\text{960}\text{W}\end{array}$

Conclusion:

Thus, the value of $V_x$ is $\underset{\_}{-100\text{V}}$ and the power absorbed by $60\Omega$ resistor is $\underset{\_}{960\text{W}}$.