Chapter 9, Problem 84P

The ac bridge shown in Fig. 9.84 is known as a Maxwell bridge and is used for accurate measurement of inductance and resistance of a coil in terms of a standard capacitance C_s. Show that when the bridge is balanced,

L_x = R₂R₃C_s and $R_x=\frac{R_2}{R_1}{R}_3$

Find L_x and R_x for R₁ = 40 kΩ, R₂ = 1.6 kΩ, R₃ = 4 kΩ, and C_s = 0.45 μF.

Figure 9.84

To determine

Show that when the bridge in Figure 9.85 is balanced the value of resistor $R_x$ is $\frac{R_2}{R_1}{R}_3$, the value of inductor $L_x$ is $R_2{R}_3{C}_s$ and also find the numerical value of $R_x$ and $L_x$.

Answer to Problem 84P

The value of resistor $R_x$ is $160\Omega$ and the value of inductor $L_x$ is $2.88\text{H}$.

Explanation of Solution

Given data:

Refer to Figure 9.85 in the textbook.

The value of resistor $\left(R_1\right)$ is $40\text{k}\Omega$.

The value of resistor $\left(R_2\right)$ is $1.6\text{k}\Omega$.

The value of resistor $\left(R_3\right)$ is $4\text{k}\Omega$.

The value of capacitor $\left(C_s\right)$ is $0.45\mu \text{F}$.

Formula used:

Write a general expression to calculate the impedance of a resistor.

$Z_R=R$        (1)

Here,

$R$ is the value of resistance.

Write a general expression to calculate the impedance of an inductor.

$Z_L=j\omega L$        (2)

Here,

$L$ is the value of inductance, and

$\omega$ is the angular frequency.

Write a general expression to calculate the impedance of a capacitor.

$Z_C=\frac{1}{j\omega C}$        (3)

Here,

$C$ is the value of capacitance.

Calculation:

The given circuit is redrawn as shown in Figure 1.

Use equation (1) to find $Z_{R_1}$, $Z_{R_2}$, $Z_{R_3}$ and $Z_{R_x}$.

$Z_{R_1}=R_1$

$Z_{R_2}=R_2$

$Z_{R_3}=R_3$

$Z_{R_x}=R_x$

Use equation (2) to find $Z_{L_x}$.

$Z_{L_x}=j\omega L_x$

Use equation (3) to find $Z_{C_s}$.

$Z_{C_s}=\frac{1}{j\omega C_s}$

Now, the impedance diagram of Figure 1 is drawn as shown in Figure 2.

Refer to Figure 2, the impedance of resistor $R_x$ and the inductor $L_x$ are connected in series.

Therefore, the equivalent impedance $\left(Z_x\right)$ is calculated as follows:

$Z_x=R_x+j\omega L_x$

Refer to Figure 2, the impedance of resistor $R_1$ and the capacitor $C_s$ are connected in parallel.

The equivalent impedance $\left(Z_1\right)$ is calculated as follows:

$\begin{array}{c}{Z}_1=R_1\parallel \frac{1}{j\omega C_s}\\ =\frac{R_1\left(\frac{1}{j\omega C_s}\right)}{R_1+\frac{1}{j\omega C_s}}\\ =\frac{R_1\left(\frac{1}{j\omega C_s}\right)}{\frac{j\omega R_1{C}_s+1}{j\omega C_s}}\\ =\frac{R_1}{j\omega R_1{C}_s+1}\end{array}$

Let,

$Z_2=Z_{R_2}=R_2$

$Z_3=Z_{R_3}=R_3$

The balance of equation of an ac bridge is,

$Z_x=\frac{Z_3}{Z_1}{Z}_2$

Substitute $\frac{R_1}{j\omega R_1{C}_s+1}$ for $Z_1$, $R_2$ for $Z_2$, $R_3$ for $Z_3$, and $R_x+j\omega L_x$ for $Z_x$ in above equation.

$\begin{array}{l}{R}_x+j\omega L_x=\left(\frac{R_3}{\left(\frac{R_1}{j\omega R_1{C}_s+1}\right)}\right)R_2\\ R_x+j\omega L_x=\frac{R_2{R}_3}{R_1}\left(1+j\omega R_1{C}_s\right)\\ R_x+j\omega L_x=\frac{R_2{R}_3}{R_1}+j\omega R_2{R}_3{C}_s\end{array}$$\frac{-j{R}_4{R}_1}{\left(\omega R_4{C}_4-j\right)}=R_3\left(R_2-\frac{j}{\omega C_2}\right)$

Equate the real and imaginary part in above equation.

$R_x=\frac{R_2{R}_3}{R_1}$        (4)

$\omega L_x=\omega R_2{R}_3{C}_s$        (5)

Simplify the equation (5) to find $L_x$.

$L_x=\frac{\omega R_2{R}_3{C}_s}{\omega }$

$L_x=R_2{R}_3{C}_s$        (6)

Substitute $40\text{k}\Omega$ for $R_1$, $1.6\text{k}\Omega$ for $R_2$, and $4\text{k}\Omega$ for $R_3$ in equation (4) to find $R_x$.

$\begin{array}{c}{R}_x=\frac{\left(1.6\text{k}\Omega \right)\left(4\text{k}\Omega \right)}{40\text{k}\Omega }\\ =\frac{\left(1.6\text{k}\Omega \right)\left(4\right)}{40}\\ =0.16\text{k}\Omega \\ =0.16\times {10}^3\Omega \left\{\because 1\text{k}={10}^3\right\}\end{array}$

$R_x=160\Omega$

Substitute $1.6\text{k}\Omega$ for $R_2$, $4\text{k}\Omega$ for $R_3$ and $0.45\mu \text{F}$ for $C$ in equation (6) to find $L_x$.

$\begin{array}{c}{L}_x=\left(1.6\text{k}\Omega \right)\left(4\text{k}\Omega \right)\left(0.45\mu \text{F}\right)\\ =\left(1.6\times {10}^3\Omega \right)\left(4\times {10}^3\Omega \right)\left(0.45\times {10}^{-6}\text{F}\right)\left\{\begin{array}{l}\because 1\text{k}={10}^3\\ 1\mu ={10}^{-6}\end{array}\right\}\\ =\left(1.6\times {10}^3\Omega \right)\left(4\times {10}^3\Omega \right)\left(0.45\times {10}^{-6}\frac{\text{s}}{\Omega }\right)\left\{\because 1\text{F}=\frac{1\text{s}}{1\Omega }\right\}\\ =2.88\Omega \cdot \text{s}\end{array}$

$L_x=2.88\text{H}\left\{\because 1\text{H}=1\Omega \cdot 1\text{s}\right\}$

Conclusion:

Thus, when the bridge is balanced, the value of resistor $R_x$ is $\frac{R_2}{R_1}{R}_3$, the value of inductor $L_x$ is $R_2{R}_3{C}_s$. The value of resistor $R_x$ is $160\Omega$ and the value of inductor $L_x$ is $2.88\text{H}$.