Chapter 4, Problem 40P

Find the Thevenin equivalent at terminals a-b of the circuit in Fig. 4.107.

Figure 4.107

To determine

Find the Thevenin voltage and Thevenin resistance at terminals a-b of the circuit shown in Figure 4.107.

Answer to Problem 40P

The Thevenin voltage is $60\text{V}$ and the Thevenin resistance is $2.857\text{kΩ}$ for the given circuit.

Explanation of Solution

Given data:

Refer to Figure 4.107 in the textbook.

The voltage source is $70\text{V}$.

Calculation:

The given circuit is modified as shown in Figure 1.

In Figure 1, apply Kirchhoff’s voltage law to the outer loop as follows,

$-70+\left(10\text{k}+20\text{k}\right)I+4V_o=0$ (1)

In Figure 1, the voltage $V_o$ across the $10\text{kΩ}$ resistor is,

$V_o=\left(10\text{k}\right)I$ (2)

Substitute equation (2) in equation (1),

$\begin{array}{l}-70+\left(10\text{k}+20\text{k}\right)I+4\left(10\text{k}\right)I=0\\ -70+\left(30\text{k}\right)I+\left(40\text{k}\right)I=0\\ -70+\left(70\text{k}\right)I=0\\ \left(70\text{k}\right)I=70\end{array}$

Simplify the equation as follows,

$\begin{array}{c}I=\frac{70}{\left(70\text{k}\right)}\\ =\frac{70}{\left(70\times {10}^3\right)}\left\{\because 1\text{k}={10}^3\right\}\\ =1\times {10}^{-3}\text{A}\\ =\text{1}\text{mA}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

In Figure 1, apply Kirchhoff’s voltage law to the Loop 1 as follows,

$-70+\left(10\text{k}\right)I+V_{\text{Th}}=0$ (3)

Substitute $1\text{mA}$ for I in equation (3) to find the Thevenin voltage in volts.

$\begin{array}{l}-70+\left(10\text{k}\right)\left(1\text{m}\right)+V_{\text{Th}}=0\\ -70+\left(10\times {10}^3\right)\left(1\times {10}^{-3}\right)+V_{\text{Th}}=0\left\{\because 1\text{m}={10}^{-3},1\text{k}={10}^3\right\}\\ -70+10+V_{\text{Th}}=0\\ V_{\text{Th}}=60\text{V}\end{array}$

In the given circuit, find the Thevenin resistance by turning off the $70\text{V}$ voltage source (replacing it with a short circuit) and connect a $1\text{V}$ source at terminals a-b.

The modified circuit is shown in Figure 2.

In Figure 2, apply Kirchhoff’s voltage law to the Loop 1 as follows,

$-1+\left(20\text{k}\right)I_1+4V_o=0$ (4)

In Figure 2, the voltage $V_o$ across the $10\text{kΩ}$ is $-1\text{V}$, as the $1\text{V}$ voltage source and the $10\text{kΩ}$ resistor are in parallel. That is,

$V_o=-1\text{V}$

Substitute $-1$ for $V_o$ in equation (4) to find the current $I_1$ in amperes.

$\begin{array}{l}-1+\left(20\text{k}\right)I_1+4\left(-1\right)=0\\ -5+\left(20\times {10}^3\right)I_1=0\left\{\because 1\text{k}={10}^3\right\}\\ \left(20\times {10}^3\right)I_1=5\\ I_1=\frac{5}{\left(20\times {10}^3\right)}\end{array}$

Simplify the equation as follows,

$\begin{array}{l}{I}_1=0.25\times {10}^{-3}\\ I_1=\text{0}\text{.25}\text{mA}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

In Figure 2, the current $I_2$ is,

$I_2=I_1+\frac{1\text{V}}{10\text{k}\Omega }$

Substitute $\text{0}\text{.25}\text{mA}$ for $I_1$ to find the current $I_2$ in amperes.

$\begin{array}{c}{I}_2=\text{0}\text{.25}\text{mA}+\frac{1\text{V}}{10\text{k}\Omega }\\ =\left(\text{0}\text{.25}\times {10}^{-3}\right)\text{A}+\left(0.1\times {10}^{-3}\right)\text{A}\left\{\because 1\text{A}=\frac{1\text{V}}{1\Omega },1\text{k}={10}^3,1\text{m}={10}^{-3}\right\}\\ =\left(3.5\times {10}^{-4}\times {10}^{-1}\times {10}^1\right)\text{A}\\ =\left(3.5\times {10}^{-1}\times {10}^{-3}\right)\text{A}\end{array}$

Simplify the equation as follows,

$\begin{array}{c}{I}_2=\left(0.35\times {10}^{-3}\right)\text{A}\\ =\text{0}\text{.35}\text{mA}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

In Figure 2, the Thevenin resistance is,

$R_{\text{Th}}=\frac{1\text{V}}{I_2}$

Substitute $\text{0}\text{.35}\text{mA}$ for $I_2$ to find the Thevenin resistance in ohms.

$\begin{array}{c}{R}_{\text{Th}}=\frac{1\text{V}}{\text{0}\text{.35}\text{mA}}\\ =\left(\frac{1}{\text{0}\text{.35}\times {\text{10}}^{-3}}\right)\Omega \left\{\because 1\Omega =\frac{1\text{V}}{1\text{A}}\right\}\\ =\left(2.857\times {\text{10}}^3\right)\Omega \\ =2.857\text{k}\Omega \left\{\because 1\text{k}={10}^3\right\}\end{array}$

Conclusion:

Thus, the Thevenin voltage is $60\text{V}$ and the Thevenin resistance is $2.857\text{kΩ}$ for the given circuit.