Chapter 4, Problem 68P

Consider the 30-Ω resistor in Fig. 4.134. First compute the Thevenin equivalent circuit as seen by the 30-Ω resistor. Compute the value of R that results in Thevenin equivalent resistance equal to the 30-Ω resistance and then calculate power delivered to the 30-Ω resistor. Now let R = 0 Ω, 110 Ω, and ∞, calculate the power delivered to the 30-Ω resistor in each case. What can you say about the value of R that will result in the maximum power that can be delivered to the 30-Ω resistor?

Figure 4.134

To determine

Find the Thevenin equivalent seen by the $30\Omega$ resistor of the circuit shown in Figure $4.134$ and find the value of R in the circuit.

Calculate the power delivered to the $30\Omega$ resistor when $R=0\Omega$, $100\Omega$, and $\infty$. Examine about the value of R that result in delivering maximum power to the $30\Omega$ resistor.

Answer to Problem 68P

The Thevenin voltage is $60\text{V}$, the Thevenin resistance is $30\Omega$, and the value of R is $50\Omega$ for the given circuit.

The power delivered to the $30\Omega$ resistor when $R=0\Omega$, $R=100\Omega$, and $R=\infty$ is $13.329\text{W}$, $61.22\text{W}$, and $120\text{W}$ respectively. The maximum power will be delivered to the $30\Omega$ resistor must be when $R=\infty$.

Explanation of Solution

Given data:

Refer to Figure 4.134 in the textbook.

The current source is $3\text{A}$.

The voltage source is $30\text{V}$.

The Thevenin resistance $R_{\text{Th}}$ is $30\Omega$.

Calculation:

In the given circuit, find the Thevenin voltage by removing the 30 ohms resistor and the modified circuit is shown in Figure 1.

The modified circuit is shown in Figure 1.

In Figure 1, the current source with parallel resistance is converted into voltage source with series resistance using source transformation. The voltage V is calculated by using ohms law as follows,

$V=3R$

The source transformation is shown in Figure 2.

In Figure 2, the Thevenin voltage is,

$V_{\text{Th}}=60\left[\frac{\left(3R-30\right)}{\left(R+10+60\right)}\right]$        (1)

Refer to Figure 4.134 in the textbook.

In the given circuit, find the Thevenin resistance by turning off the $30\text{V}$ voltage (replace with a short circuit) and $3\text{A}$ current source (replace with an open circuit).

The modified circuit is shown in Figure 3.

In Figure 3, the Thevenin resistance is,

$R_{\text{Th}}=60\left|\right|\left(R+10\right)$

$R_{\text{Th}}=\frac{60\left(R+10\right)}{R+60+10}$        (2)

Substitute $30$ for $R_{\text{Th}}$ in equation (2) to find the value of the resistor R in ohms.

$\begin{array}{l}30=\frac{60\left(R+10\right)}{\left(R+60+10\right)}\\ 30\left(R+60+10\right)=60\left(R+10\right)\\ 30R+1800+300=60R+600\\ R=50\Omega \end{array}$

Substitute 50 for R in equation (1) to find the Thevenin voltage in volts.

$\begin{array}{c}{V}_{\text{Th}}=60\left[\frac{\left(3\left(50\right)-30\right)}{\left(50+10+60\right)}\right]\\ =\frac{60\times 120}{120}\\ =60\text{V}\end{array}$

Substitute 50 for R in equation (2) to find the Thevenin resistance in ohms.

$\begin{array}{c}{R}_{\text{Th}}=\frac{60\left(50+10\right)}{50+60+10}\\ =30\Omega \end{array}$

The Thevenin equivalent connected to the 30 ohms resistor is shown in Figure 4.

The power delivered to the 30 ohms resistor is,

$p_{30\Omega }={\left[\frac{V_{\text{Th}}}{R_{\text{Th}}+30}\right]}^2\left(30\right)$        (3)

Substitute 60 for $V_{\text{Th}}$ and 30 for $R_{\text{Th}}$ in equation (3) to find the power delivered to the 30 ohms resistor in watts.

$\begin{array}{c}{p}_{30\Omega }={\left[\frac{60}{30+30}\right]}^2\left(30\right)\\ =30\text{W}\end{array}$

Consider the resistance $R=0\Omega$.

When $R=0\Omega$, the Thevenin resistance is calculated as follows.

Substitute 0 for R in equation (2) to find the Thevenin resistance in ohms.

$\begin{array}{c}{R}_{\text{Th}}=\frac{60\left(0+10\right)}{0+60+10}\\ =8.571\Omega \end{array}$

When $R=0\Omega$, the Thevenin voltage is calculated as follows.

Substitute 0 for R in equation (1) to find the Thevenin voltage in volts.

$\begin{array}{c}{V}_{\text{Th}}=60\left[\frac{\left(3\left(0\right)-30\right)}{\left(0+10+60\right)}\right]\\ =-25.71\text{V}\end{array}$

When $R=0\Omega$, the power delivered to the 30 ohms resistor is calculated as follows.

Substitute $-25.71$ for $V_{\text{Th}}$ and 8.571 for $R_{\text{Th}}$ in equation (3) to find the power delivered to the 30 ohms resistor in watts.

$\begin{array}{c}{p}_{30\Omega }={\left[\frac{-25.71}{8.571+30}\right]}^2\left(30\right)\\ =13.329\text{W}\end{array}$

Consider the resistance $R=110\Omega$.

When $R=110\Omega$, the Thevenin resistance is calculated as follows.

Substitute 110 for R in equation (2) to find the Thevenin resistance in ohms.

$\begin{array}{c}{R}_{\text{Th}}=\frac{60\left(110+10\right)}{110+60+10}\\ =40\Omega \end{array}$

When $R=110\Omega$, the Thevenin voltage is calculated as follows.

Substitute 110 for R in equation (1) to find the Thevenin voltage in volts.

$\begin{array}{c}{V}_{\text{Th}}=60\left[\frac{\left(3\left(110\right)-30\right)}{\left(110+10+60\right)}\right]\\ =100\text{V}\end{array}$

When $R=110\Omega$, the power delivered to the 30 ohms resistor is calculated as follows.

Substitute $100$ for $V_{\text{Th}}$ and 40 for $R_{\text{Th}}$ in equation (3) to find the power delivered to the 30 ohms resistor in watts.

$\begin{array}{c}{p}_{30\Omega }={\left[\frac{100}{40+30}\right]}^2\left(30\right)\\ =61.22\text{W}\end{array}$

Consider the resistance $R=\infty$.

When $R=\infty$, the Thevenin resistance is calculated as follows.

Simplify equation (2) as follows,

$\begin{array}{c}{R}_{\text{Th}}=\frac{60R\left(1+\frac{10}{R}\right)}{R\left(1+\frac{60}{R}+\frac{10}{R}\right)}\\ =\frac{60\left(1+\frac{10}{R}\right)}{1+\frac{60}{R}+\frac{10}{R}}\end{array}$

Substitute $\infty$ for R in to find the Thevenin resistance in ohms.

$\begin{array}{c}{R}_{\text{Th}}=\frac{60\left(1+\frac{10}{\infty }\right)}{1+\frac{60}{\infty }+\frac{10}{\infty }}\\ =\frac{60\left(1+0\right)}{1+0+0}\left\{\because \frac{1}{\infty }=0\right\}\\ =60\Omega \end{array}$

When $R=\infty$, the Thevenin voltage is calculated as follows.

Simplify equation (1) as follows,

$\begin{array}{c}{V}_{\text{Th}}=60\left[\frac{\left(3R-30\right)}{\left(R+10+60\right)}\right]\\ =60\left[\frac{R\left(3-\frac{30}{R}\right)}{R\left(1+\frac{10}{R}+\frac{60}{R}\right)}\right]\\ =60\left[\frac{3-\frac{30}{R}}{1+\frac{10}{R}+\frac{60}{R}}\right]\end{array}$

Substitute $\infty$ for R to find the Thevenin voltage in volts.

$\begin{array}{c}{V}_{\text{Th}}=60\left[\frac{3-\frac{30}{\infty }}{1+\frac{10}{\infty }+\frac{60}{\infty }}\right]\\ =60\left[\frac{3-0}{1+0+0}\right]\left\{\because \frac{1}{\infty }=0\right\}\\ =180\text{V}\end{array}$

When $R=\infty$, the power delivered to the 30 ohms resistor is calculated as follows.

Substitute $180$ for $V_{\text{Th}}$ and 60 for $R_{\text{Th}}$ in equation (3) to find the power delivered to the 30 ohms resistor in watts.

$\begin{array}{c}{p}_{30\Omega }={\left[\frac{180}{60+30}\right]}^2\left(30\right)\\ =120\text{W}\end{array}$

Thus, when $R=\infty$, the maximum power will be delivered to the 30 ohms resistor.

Conclusion:

Thus, the Thevenin voltage is $60\text{V}$, the Thevenin resistance is $30\Omega$, and the value of R is $50\Omega$ for the given circuit.

The power delivered to the $30\Omega$ resistor when $R=0\Omega$, $R=100\Omega$, and $R=\infty$ is $13.329\text{W}$, $61.22\text{W}$, and $120\text{W}$ respectively. The maximum power will be delivered to the $30\Omega$ resistor must be when $R=\infty$.