Chapter 4, Problem 92P

Consider the bridge circuit of Fig. 4.148. Is the bridge balanced? If the 10-kΩ resistor is replaced by an 18-kΩ resistor, what resistor connected between terminals a-b absorbs the maximum power? What is this power?

Figure 4.148

To determine

Find whether the bridge circuit of Figure 4.148 is balanced. Also find the value of the load resistor connected between the terminals a-b of the circuit and the maximum power absorbed by the load resistor.

Answer to Problem 92P

The bridge circuit is balanced when $\text{10}\text{k}\Omega$ resistor is used and unbalanced when $\text{18}\text{k}\Omega$ resistor is used. The value of the load resistor at terminals a-b is $6.398\text{k}\Omega$ and the maximum power absorbed by the load resistor is $16.622\text{mW}$.

Explanation of Solution

Given data:

Refer to Figure 4.148 in the textbook.

The voltage source is $220\text{V}$.

Calculation:

For a bridge to be balanced, the voltage measured at terminals a-b should be zero.

The given circuit is modified as shown in Figure 1.

In Figure 1, apply Kirchhoff’s voltage law at loop $i_1$ as follows.

$\begin{array}{l}2i_1+8\left(i_1-i_2\right)=220\\ 2i_1+8i_1-8i_2=220\end{array}$

$10i_1-8i_2=220$        (1)

In Figure 1, apply Kirchhoff’s voltage law at loop $i_2$ as follows.

$\begin{array}{l}24i_2-8i_1=0\\ 24i_2=8i_1\\ i_2=\frac{8i_1}{24}\end{array}$

$i_2=\left(\frac{1}{3}\right)i_1$        (2)

Substitute equation (2) in equation (1) as follows,

$\begin{array}{l}10i_1-8\left(\frac{1}{3}\right)i_1=220\\ 30i_1-8i_1=660\\ 22i_1=660\\ i_1=\frac{660}{22\text{k}}\end{array}$

Simplify the equation as follows,

$\begin{array}{c}{i}_1=\frac{660}{22\times {10}^3}\left\{\because 1\text{k}={10}^3\right\}\\ =30\times {10}^{-3}\text{A}\\ =\text{30}\text{mA}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

Substitute $30\text{m}$ for $i_1$ in equation (2) to find the current $i_2$ in amperes.

$\begin{array}{c}{i}_2=\left(\frac{1}{3}\right)\left(30\text{m}\right)\\ =10\text{mA}\end{array}$

In Figure 1, apply Kirchhoff’s voltage law at loop 1 as follows.

$v_{ab}+5\left(i_2-i_1\right)+10i_2=0$        (3)

Substitute $30\text{m}$ for $i_1$ and $10\text{m}$ for $i_2$ in equation (3) to find the voltage $v_{ab}$ in volts.

$\begin{array}{l}{v}_{ab}+5\left(10\text{m}-30\text{m}\right)+10\left(10\text{m}\right)=0\\ v_{ab}-100\text{m}+\text{100}\text{m}=0\\ v_{ab}=0\end{array}$

As the voltage $v_{ab}$ at terminals a-b is zero, the bridge is balanced.

In Figure 1, when the $10\text{k}\Omega$ resistor is replaced with the $18\text{k}\Omega$ resistor, the bridge becomes unbalanced and the modified Figure is shown in Figure 2.

In Figure 2, apply Kirchhoff’s voltage law at loop $i_1$ and remains the same as given in equation (1). Therefore,

$10i_1-8i_2=220$        (4)

In Figure 2, apply Kirchhoff’s voltage law at loop $i_2$ as follows.

$\begin{array}{l}32i_2-8i_1=0\\ 32i_2=8i_1\\ i_2=\frac{8i_1}{32}\end{array}$

$i_2=\left(\frac{1}{4}\right)i_1$        (5)

Substitute $\left(\frac{1}{4}\right)i_1$ for $i_2$ in equation (4) as follows.

$\begin{array}{l}10i_1-8\left(\frac{1}{4}\right)i_1=220\\ 40i_1-8i_1=880\\ i_1=\frac{880}{32\text{k}}\end{array}$

Simplify the equation as follows,

$\begin{array}{c}{i}_1=\frac{880}{32\times {10}^3}\left\{\because 1\text{k}={10}^3\right\}\\ =27.5\times {10}^{-3}\text{A}\\ =27.5\text{mA}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

Substitute $27.5\text{m}$ for $i_1$ in equation (5) to find the current $i_2$ in amperes.

$\begin{array}{c}{i}_2=\left(\frac{1}{4}\right)\left(27.5\text{m}\right)\\ =6.875\text{mA}\end{array}$

In Figure 2, the voltage $v_{ab}$ at terminals a-b is,

$v_{ab}=5\left(i_1-i_2\right)-18i_2$

Substitute $27.5$ for $i_1$ and $6.875$ for $i_2$ to find the voltage $v_{ab}$ in volts.

$\begin{array}{c}{v}_{ab}=5\left(27.5-6.875\right)-18\left(6.875\right)\\ =\left(137.5\right)-\left(34.375\right)-\left(123.75\right)\\ =-20.625\text{V}\end{array}$

Since the voltage $v_{ab}$ at terminals a-b is the Thevenin voltage. Therefore,

$V_{\text{Th}}=v_{ab}=-20.625\text{V}$

Refer to Figure (2).

In Figure (2), find the Thevenin resistance by turning off the $220\text{V}$ voltage source (replacing with an open circuit) and the modified circuit is shown in Figure 3.

In Figure 3, $2\text{k}\Omega$, $3\text{k}\Omega$, and $5\text{k}\Omega$ resistors are in delta connection. This delta connected resistors is connected to the wye connection as shown in Figure 4.

For the wye connection in Figure 4, the value of the resistor $R_1$ is calculated as follows,

$\begin{array}{c}{R}_1=\frac{\left(3\text{k}\times 5\text{k}\right)}{\left(2\text{k}+3\text{k}\times 5\text{k}\right)}\\ =\frac{15{\text{k}}^2}{10\text{k}}\\ =1.5\text{k}\Omega \end{array}$

For the wye connection in Figure 4, the value of the resistor $R_2$ is calculated as follows,

$\begin{array}{c}{R}_2=\frac{\left(2\text{k}\times 3\text{k}\right)}{\left(2\text{k}+3\text{k}\times 5\text{k}\right)}\\ =\frac{6{\text{k}}^2}{10\text{k}}\\ =0.6\text{k}\Omega \end{array}$

For the wye connection in Figure 4, the value of the resistor $R_3$ is calculated as follows,

$\begin{array}{c}{R}_3=\frac{\left(2\text{k}\times 5\text{k}\right)}{\left(2\text{k}+3\text{k}\times 5\text{k}\right)}\\ =\frac{10{\text{k}}^2}{10\text{k}}\\ =1\text{k}\Omega \end{array}$

Figure 4 is modified as shown in Figure 5.

In Figure 5, the Thevenin resistance is,

$\begin{array}{c}{R}_{\text{Th}}=\left(1.5\text{k}\right)\text{+}\left(0.6\text{k}+6\text{k}\right)\left|\right|\left(1\text{k+18}\text{k}\right)\\ =\left(1.5\text{k}\right)+\left(6.6\text{k||19}\text{k}\right)\\ =\left(1.5\text{k}\right)+\left(\frac{6.6\text{k}\times \text{19}\text{k}}{6.6\text{k+19}\text{k}}\right)\\ =1.5\text{k}+4.898\text{k}\end{array}$

Simplify the equation as follows,

$R_{\text{Th}}=6.398\text{k}\Omega$

For maximum power transfer,

$R_L=R_{\text{Th}}=6.398\text{k}\Omega$

The maximum power absorbed by the load resistor is,

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

Substitute $-20.625$ for $V_{\text{Th}}$ and $6.398\text{k}$ for $R_{\text{Th}}$ to find the maximum power absorbed by the load resistor in ohms.

$\begin{array}{c}{p}_{\max }=\frac{{\left(-20.625\right)}^2}{4\left(6.398\times {10}^3\right)}\left\{\because 1\text{k}={10}^3\right\}\\ =16.622\times {10}^{-3}\\ =16.622\text{mW}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

Conclusion:

Thus, the bridge circuit is balanced when $\text{10}\text{k}\Omega$ resistor is used and unbalanced when $\text{18}\text{k}\Omega$ resistor is used. The value of the load resistor at terminals a-b is $6.398\text{k}\Omega$ and the maximum power absorbed by the load resistor is $16.622\text{mW}$.