The arms of an a.c. Maxwell bridge are arranged as follows:

AB is a non-inductive resistance of 2,000 Ω in parallel with a capacitor of capacitance 0.5 μF, BC is a non-inductive resistance of 500 Ω, CD is an inductive impedance (unknown), and DA is a non-inductive resistance of 400 Ω. If balance is obtained under these conditions, find the value of the Q factor of the branch CD.

Concept:

At bridge balance condition,

\(\left( {{R_1} + j\omega {L_1}} \right)\left( {\frac{{\left( {{R_3}} \right)\left( {\frac{1}{{j\omega {C_3}}}} \right)}}{{{R_3} + \frac{1}{{j\omega {C_3}}}}}} \right) = {R_2}{R_4}\)

\(\left( {{R_1} + j\omega {L_1}} \right)\left( {\frac{{{R_3}}}{{1 + j\omega {R_3}{C_3}}}} \right) = {R_2}{R_4}\)

\(\left( {{R_1} + j\omega {L_1}} \right){R_3} = {R_2}{R_4}\left( {1 + j\omega {R_3}{C_3}} \right)\)

\( \Rightarrow {R_1}{R_3} + j\omega {L_1}{R_3} = {R_2}{R_4} + j\omega {R_3}{C_3}{R_2}{R_4}\)

By comparing both the sides,

\({R_1} = \frac{{{R_2}{R_4}}}{{{R_3}}} = \frac{{500 \times 400}}{{2000 }} = 100\;{\rm{\Omega }}\)

\({L_1} = {C_3}{R_2}{R_4} = 0.5 \times {10^{ - 6}} \times 500 \times 400 = 0.1\;H\)

The quality factor, \(Q = \frac{{\omega {L_1}}}{{{R_1}}}\) 

\(= \frac{{2\pi \times 50 \times 0.1}}{{100}} = 0.3141\)