Question:

Published on: 1 October, 2023

**Explain the worming of Anderson’s bridge with a neat sketch. Derive the required expression for obtaining the unknown inductance. **

Answer:

By using this bridge, the unknown inductance is measured in terms of a known capacitance and resistance.

Let, \(L_1\) = Self-inductance to be measured

\(R_1\) = Resistance of self-inductor

\(C\) = Fixed Capacitor

\(R_1\) = Resistance connected in series with self-inductor

\(R_4, R_3, R_2, r\) = Known non-inductive resistances.

At balance, \(I_1 = I_3\) and \(I_2 = I_C+I_4\)

\(I_1R_3=I_C\times\frac{1}{j\omega C}\)

\(I_C=I_1R_3j\omega C\)

Writing other balance equations,

\(I_1\left(r_1+R_1+j\omega L_1\right)=I_2R_2+I_Cr\) ........(1)

and, \(I_C\left(r+\frac{1}{j\omega C}\right)=\left(I_2-I_C\right)R_4\) .........(2)

Putting the value of \({I}_C\) in equation \(\left(1\right)\), we get,

\(I_1\left(r_1+R_1+j\omega L_1\right)=I_2R_2+I_1R_3j\omega Cr\)

\(I_1\left(r_1+R_1+j\omega L_1-R_3j\omega Cr\right)=I_2R_2\) ........(3)

Putting the value of \(I_C\) in equation (2)

\(I_1R_3j\omega C\left(r+\frac{1}{j\omega C}\right)=\left(I_2-I_1R_3j\omega C\right)R_4\)

\(I_1\left(R_3j\omega Cr+R_3j\omega C R_4+R_3\ \right)=I_2R_4\) .........(4)

From equation (3) & (4) we get,

\(I_1\left(r_1+R_1+j\omega L_1-R_3j\omega Cr\right)=I_1\left(\frac{R_2R_3}{R_4}+\frac{j\omega C R_2R_3r}{R_4}+j\omega C R_2R_3\right)\)

Equating the real and imaginary parts \(R_1=\frac{R_2R_3}{R_4}-r_1\)

\(L_1=c\frac{R_3}{R_4}[r\left(\ R_4+R_2\right)+R_2R_4]\)

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