Chapter 9, Problem 85P

The ac bridge circuit of Fig. 9.85 is called a Wien bridge. It is used for measuring the frequency of a source. Show that when the bridge is balanced,

$f=\frac{1}{2\pi \sqrt{R_2{R}_4{C}_2{C}_4}}$

Figure 9.85

To determine

Show that when the bridge in Figure 9.85 is balanced the value of frequency is $f=\frac{1}{2\pi \sqrt{R_2{R}_4{C}_2{C}_4}}$.

Explanation of Solution

Given data:

Refer to Figure 9.85 in the textbook.

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 a capacitor.

$Z_C=\frac{1}{j\omega C}$        (2)

Here,

$C$ is the value of capacitance.

Write a general expression to calculate the angular frequency.

$\omega =2\pi f$        (3)

Here,

$f$ is the value of frequency.

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 $Z_{R_4}$.

$Z_{R_1}=R_1$

$Z_{R_2}=R_2$

$Z_{R_3}=R_3$

$Z_{R_4}=R_4$

Use equation (2) to find $Z_{C_2}$ and $Z_{C_4}$.

$Z_{C_2}=\frac{1}{j\omega C_2}$

$Z_{C_4}=\frac{1}{j\omega C_4}$

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

Refer to Figure 2, the impedance of resistor $R_2$ and the capacitor $C_2$ are connected in series.

Therefore, the equivalent impedance $\left(Z_2\right)$ is calculated as follows:

$\begin{array}{c}{Z}_2=R_2+\frac{1}{j\omega C_2}\\ =R_2-\frac{j}{\omega C_2}\end{array}$

Refer to Figure 2, the impedance of resistor $R_4$ and the capacitor $C_4$ are connected in parallel.

The equivalent impedance $\left(Z_4\right)$ is calculated as follows:

$\begin{array}{c}{Z}_4=R_4\parallel \frac{1}{j\omega C_4}\\ =\frac{R_4\left(\frac{1}{j\omega C_4}\right)}{R_4+\frac{1}{j\omega C_4}}\\ =\frac{R_4\left(\frac{1}{j\omega C_4}\right)}{\frac{j\omega R_4{C}_4+1}{j\omega C_4}}\\ =\frac{R_4}{j\omega R_4{C}_4+1}\end{array}$

Simplify the above equation as follows:

$\begin{array}{c}{Z}_4=\frac{R_4}{j\left(\omega R_4{C}_4+\frac{1}{j}\right)}\\ =\frac{-j{R}_4}{\left(\omega R_4{C}_4-j\right)}\end{array}$

Let,

$Z_1=Z_{R_1}=R_1$

$Z_3=Z_{R_3}=R_3$

The balance of equation of an ac bridge is,

$Z_4=\frac{Z_3}{Z_1}{Z}_2$

The above equation as rearranged as follows:

$Z_1{Z}_4=Z_2{Z}_3$

Substitute $R_1$ for $Z_1$, $R_2-\frac{j}{\omega C_2}$ for $Z_2$, $R_3$ for $Z_3$, and $\frac{-j{R}_4}{\left(\omega R_4{C}_4-j\right)}$ for $Z_4$ in above equation.

$\frac{-j{R}_4{R}_1}{\left(\omega R_4{C}_4-j\right)}=R_3\left(R_2-\frac{j}{\omega C_2}\right)$

Take the conjugate of denominator to rationalize the fraction in left hand side.

$\begin{array}{l}\frac{-j{R}_4{R}_1\left(\omega R_4{C}_4+j\right)}{\left(\omega R_4{C}_4-j\right)\left(\omega R_4{C}_4+j\right)}=R_3{R}_2-\frac{j{R}_3}{\omega C_2}\\ \frac{\left(-j\omega R_1{R}_4^2{C}_4-j^2{R}_4{R}_1\right)}{{\left(\omega R_4{C}_4\right)}^2-j^2}=R_3{R}_2-\frac{j{R}_3}{\omega C_2}\\ \frac{\left(-j\omega R_1{R}_4^2{C}_4-\left(-1\right)R_4{R}_1\right)}{{\left(\omega R_4{C}_4\right)}^2-{\left(-1\right)}^2}=R_3{R}_2-\frac{j{R}_3}{\omega C_2}\left\{\because j^2=-1\right\}\\ \frac{\left(R_1{R}_4-j\omega R_1{R}_4^2{C}_4\right)}{{\omega }^2{R}_4^2{C}_4^2+1}=R_3{R}_2-\frac{j{R}_3}{\omega C_2}\end{array}$

$\frac{R_1{R}_4}{{\omega }^2{R}_4^2{C}_4^2+1}-\frac{j\omega R_1{R}_4^2{C}_4}{{\omega }^2{R}_4^2{C}_4^2+1}=R_3{R}_2-\frac{j{R}_3}{\omega C_2}$

Equate the real and imaginary part in above equation.

$\frac{R_1{R}_4}{{\omega }^2{R}_4^2{C}_4^2+1}=R_3{R}_2$        (4)

$-\frac{\omega R_1{R}_4^2{C}_4}{{\omega }^2{R}_4^2{C}_4^2+1}=-\frac{R_3}{\omega C_2}$

$\frac{\omega R_1{R}_4^2{C}_4}{{\omega }^2{R}_4^2{C}_4^2+1}=\frac{R_3}{\omega C_2}$        (5)

Divide equation (4) by (5).

$\begin{array}{l}\frac{\left(\frac{R_1{R}_4}{{\omega }^2{R}_4^2{C}_4^2+1}\right)}{\left(\frac{\omega R_1{R}_4^2{C}_4}{{\omega }^2{R}_4^2{C}_4^2+1}\right)}=\frac{R_3{R}_2}{\left(\frac{R_3}{\omega C_2}\right)}\\ \left(\frac{R_1{R}_4}{{\omega }^2{R}_4^2{C}_4^2+1}\right)\left(\frac{{\omega }^2{R}_4^2{C}_4^2+1}{\omega R_1{R}_4^2{C}_4}\right)=\frac{\omega R_2{R}_3{C}_2}{j{R}_3}\\ \frac{1}{\omega R_4{C}_4}=\omega R_2{C}_2\\ {\omega }^2=\frac{1}{R_2{C}_2{R}_4{C}_4}\end{array}$

Simplify the above equation to find $\omega$.

$\begin{array}{c}\omega =\sqrt{\frac{1}{R_2{C}_2{R}_4{C}_4}}\\ =\frac{1}{\sqrt{R_2{C}_2{R}_4{C}_4}}\end{array}$

Rearrange equation (3) to find $f$.

$f=\frac{\omega }{2\pi }$

Substitute $\frac{1}{\sqrt{R_2{C}_2{R}_4{C}_4}}$ for $\omega$ in above equation to find $f$.

$\begin{array}{c}f=\frac{\left(\frac{1}{\sqrt{R_2{C}_2{R}_4{C}_4}}\right)}{2\pi }\\ =\frac{1}{2\pi \sqrt{R_2{C}_2{R}_4{C}_4}}\end{array}$

Conclusion:

Thus, when the bridge is balanced, the value of frequency $f=\frac{1}{2\pi \sqrt{R_2{C}_2{R}_4{C}_4}}$ is shown.