Chapter 4, Problem 69P

Find the maximum power transferred to resistor R in the circuit of Fig. 4.135.

Figure 4.135

To determine

Calculate the maximum power delivered to the resistor R of the circuit shown in Figure 4.135.

Answer to Problem 69P

The maximum power delivered to the variable resistor R is infinity.

Explanation of Solution

Given data:

Refer to Figure 4.135 in the textbook.

The voltage source is 100 V.

Formula used:

Write the expression to find the power delivered to the resistor.

$p_{\max }=\frac{V_{\text{Th}}^2}{4R_{\text{Th}}}$        (1)

Here,

$V_{\text{Th}}$ is the Thevenin voltage, and

$R_{\text{Th}}$ is the Thevenin resistance.

Calculation:

The given circuit is modified as shown in Figure 1.

In Figure 1, the voltage source with series resistance is converted into current source with parallel resistance using source transformation. The current (I) is calculated by using Ohm’s law.

$\begin{array}{c}I=\frac{100\text{V}}{10\times {10}^3\Omega }\left\{\because 1\text{k}={10}^3\right\}\\ =10\times {10}^{-3}\text{A}\left\{\because 1\text{A}=\frac{1\text{V}}{1\Omega }\right\}\\ =10\text{mA }\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

The source transformation is shown in Figure 2.

In Figure 2, apply Kirchhoff’s current law at node voltage $v_1$ as follows.

$\begin{array}{l}\left(\frac{v_1-v_2}{22\times {10}^3}\right)-\left(0.006v_2\right)+\left(\frac{v_1-0}{30\times {10}^3}\right)=0\left\{\because v_o=v_2,1\text{k}={10}^3\right\}\\ \left(0.0454545\times {10}^{-3}\right)\left(v_1-v_2\right)-\left(0.006v_2\right)+\left(0.0333\times {10}^{-3}\right)v_1=0\\ \left(0.0454545\times {10}^{-3}\right)v_1-\left(0.0454545\times {10}^{-3}\right)v_2-\left(0.006v_2\right)+\left(0.03333\times {10}^{-3}\right)v_1=0\\ \left(7.87845\times {10}^{-5}\right)v_1-\left(6.0454545\times {10}^{-3}\right)v_2=0\end{array}$

Simplify the equation as follows,

$v_1=\frac{\left(6.0454545\times {10}^{-3}\right)v_2}{\left(7.87845\times {10}^{-5}\right)}$

$v_1=76.734v_2$        (2)

In Figure 2, apply Kirchhoff’s current law at node voltage $v_2$ as follows.

$-0.01+\left(\frac{v_2-0}{10\times {10}^3}\right)+\left(\frac{v_2-0}{40\times {10}^3}\right)+\left(\frac{v_2-v_1}{22\times {10}^3}\right)=0\left\{\because 1\text{k}={10}^{-3}\right\}$        (3)

$\begin{array}{l}-0.01+\left(0.1\times {10}^{-3}\right)v_2+\left(0.025\times {10}^{-3}\right)v_2+\left(0.0454545\times {10}^{-3}\right)v_2-\left(0.0454545\times {10}^{-3}\right)v_1=0\\ \left(1.704545\times {10}^{-4}\right)v_2-\left(0.0454545\times {10}^{-3}\right)v_1=0.01\end{array}$

Substitute $76.734v_2$ for $v_1$ to find the node voltage $v_2$ in volts.

$\begin{array}{l}\left(1.704545\times {10}^{-4}\right)v_2-\left(0.0454545\times {10}^{-3}\right)\left(76.734v_2\right)=0.01\\ \left(1.704545\times {10}^{-4}\right)v_2-\left(3.4879\times {10}^{-3}\right)v_2=0.01\\ \left(-3.317445\times {10}^{-3}\right)v_2=0.01\end{array}$

Simplify the equation as follows,

$v_2=-3.0143\text{V}$

Substitute $-3.0143\text{V}$ for $v_2$ in equation (2) to find the node voltage $v_1$ in volts.

$\begin{array}{l}{v}_1=76.734\left(-3.0143\text{V}\right)\\ v_1=-231.3\text{V}\end{array}$

Since, the voltage at node $v_1$ is the open circuit voltage. Therefore,

$v_{oc}=v_1=-231.3\text{V}$

Refer to Figure 4.135 in the textbook.

In the given circuit, find the short circuit current by shorting the resistor R.

The modified circuit is shown in Figure 3.

To find the short circuit current $i_{sc}$, set $v_1=0\text{V}$.

Substitute 0 for $v_1$ in equation (3) to find the node voltage $v_2$ in volts.

$\begin{array}{l}-0.01+\left(\frac{v_2-0}{10\times {10}^3}\right)+\left(\frac{v_2-0}{40\times {10}^3}\right)+\left(\frac{v_2-0}{22\times {10}^3}\right)=0\\ -0.01+\left(0.1\times {10}^{-3}\right)v_2+\left(0.025\times {10}^{-3}\right)v_2+\left(0.0454545\times {10}^{-3}\right)v_2=0\\ \left(1.704545\times {10}^{-4}\right)v_2=0.01\\ v_2=58.6667\text{V}\end{array}$

In Figure 3, apply Kirchhoff’s current law at node voltage $v_1$ as follows.

$\begin{array}{l}\left(\frac{v_1-v_2}{22\times {10}^3}\right)-0.006v_2+i_{sc}=0\left\{\because v_o=v_2\right\}\\ i_{sc}=0.006v_2-\left(\frac{v_1-v_2}{22\times {10}^3}\right)\end{array}$

Substitute 0 for $v_1$ and $58.6667$ for $v_2$ to find the short circuit current $i_{sc}$ in amperes.

$\begin{array}{c}{i}_{sc}=0.006\left(58.6667\right)-\left(\frac{0-58.6667}{22\times {10}^3}\right)\\ =0.352+\left(2.666\times {10}^{-3}\right)\\ =0.35467\text{A}\end{array}$

The Thevenin resistance is,

$R_{\text{Th}}=\frac{v_{oc}}{i_{sc}}$

Substitute $-231.3$ for $v_{oc}$ and $0.35467$ for $i_{sc}$ to find the Thevenin resistance in ohms.

$\begin{array}{c}{R}_{\text{Th}}=\frac{-231.3}{0.35467}\\ =-652.2\Omega \end{array}$

The negative equivalent resistance indicates that an active device (dependent source) presents in the circuit, since the circuit cannot have a negative resistance in a purely passive circuit. The negative resistance for the equivalent circuit means that both the resistance and the source will effectively delivers the power to load.

Since the resistance cannot be the negative, the correct answer will be $R=652.2\Omega$ which then means the current through the resistor R is infinite and the power delivered to the resistor R is infinity.

Conclusion:

Thus, the maximum power delivered to the variable resistor R is infinity.