Question 9.84: The ac bridge shown in Fig. 9.84 is known as a Maxwell bridg...

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,

\begin{aligned}&L_{x}=R_{2} R_{3} C_{s} \quad \text { and } \quad R_{x}=\frac{R_{2}}{R_{1}} R_{3}\\\\&\text { Find } L_{x} \text { and } R_{x} \text { for } R_{1}=40 \mathrm{k} \Omega, R_{2}=1.6 \mathrm{k} \Omega, R_{3}=4 \mathrm{k} \Omega, \text { and } C_{s}=0.45 \mu \mathrm{F}\end{aligned}
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Let \quad {Z}_{1}=\mathrm{R}_{1} \| \frac{1}{\mathrm{j} \omega \mathrm{C}_{\mathrm{s}}}, \quad \mathrm{Z}_{2}=\mathrm{R}_{2}, \quad \mathrm{Z}_{3}=\mathrm{R}_{3}, and \mathrm{Z}_{\mathrm{x}}=\mathrm{R}_{\mathrm{x}}+\mathrm{j} \omega \mathrm{L}_{\mathrm{x}}\\\\

{Z}_{1}=\frac{\frac{\mathrm{R}_{1}}{\mathrm{j} \omega \mathrm{C}_{\mathrm{s}}}}{\mathrm{R}_{1}+\frac{1}{\mathrm{j} \omega \mathrm{C}_{\mathrm{s}}}}=\frac{\mathrm{R}_{1}}{\mathrm{j} \omega \mathrm{R}_{1} \mathrm{C}_{\mathrm{s}}+1}\\\\

since  {Z}_{\mathrm{x}}=\frac{{Z}_{3}}{{Z}_{1}} {Z}_{2}\\\\

\mathrm{R}_{\mathrm{x}}+\mathrm{j} \omega \mathrm{L}_{\mathrm{x}}=\mathrm{R}_{2} \mathrm{R}_{3} \frac{\mathrm{j} \omega \mathrm{R}_{1} \mathrm{C}_{\mathrm{s}}+1}{\mathrm{R}_{1}}=\frac{\mathrm{R}_{2} \mathrm{R}_{3}}{\mathrm{R}_{1}}\left(1+\mathrm{j} \omega \mathrm{R}_{1} \mathrm{C}_{\mathrm{s}}\right)\\\\

Equating the real and imaginary components,

{R}_{\mathrm{x}}=\frac{{R}_{2} {R}_{3}}{{R}_{1}}\\\\

\omega \mathrm{L}_{\mathrm{x}}=\frac{\mathrm{R}_{2} \mathrm{R}_{3}}{\mathrm{R}_{1}}\left(\omega \mathrm{R}_{1} \mathrm{C}_{\mathrm{s}}\right)\\\\ implies that

{{L}_{{x}}}={R}_{2} {R}_{3} {C}_{{s}}\\\\

Given that \mathrm{R}_{1}=40 \mathrm{k} \Omega, \mathrm{R}_{2}=1.6 \mathrm{k} \Omega, \mathrm{R}_{3}=4 \mathrm{k} \Omega, and \mathrm{C}_{\mathrm{s}}=0.45 \mu \mathrm{F}\\\\

\begin{array}{l}\mathrm{R}_{\mathrm{x}}=\frac{\mathrm{R}_{2} \mathrm{R}_{3}}{\mathrm{R}_{1}}=\frac{(1.6)(4)}{40} \mathrm{k} \Omega=0.16 \mathrm{k} \Omega={160} \Omega \\\\\mathrm{L}_{\mathrm{x}}=\mathrm{R}_{2} \mathrm{R}_{3} \mathrm{C}_{\mathrm{s}}=(1.6)(4)(0.45)={2.88} \mathrm{H}\\\end{array}

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