GATE 2022 IN Q.22: solution and answer

Official answer: A, C, DSource: Official answer key, GATE 2022 IN (IIT Kharagpur)

1. Test used: for an AC bridge with arms $Z_1,Z_2,Z_3,Z_4$ (where $Z_1$ and $Z_4$ are opposite arms), balance requires $Z_1{Z}_4=Z_2{Z}_3$. This one complex equation gives two real conditions, one from the real parts and one from the imaginary parts. The balance is frequency-independent if $\omega$ ends up in neither condition, either because it never appears or because it cancels.

2. Maxwell (inductance–capacitance) bridge: arm 1 is $R_x+j\omega L_x$, arm 2 is $R_2$, arm 3 is $R_3$ and arm 4 is $R_4\parallel C_4$, so $Y_4=\frac{1}{R_4}+j\omega C_4$. Then $$Z_1=Z_2{Z}_3{Y}_4\Rightarrow R_x+j\omega L_x=R_2{R}_3\left(\frac{1}{R_4}+j\omega C_4\right)\Rightarrow R_x=\frac{R_2{R}_3}{R_4},L_x=R_2{R}_3{C}_4$$ In the imaginary-part equation $\omega$ appears on both sides and cancels, so the balance is frequency-independent. (The Maxwell inductance bridge, which compares $L_x$ with a standard inductor $L_2$ in series with $R_2$, gives $L_x=L_2{R}_3/R_4$. It is also frequency-independent.)

3. Wien bridge: arm 1 is $R_1$ in series with $C_1$, arm 2 is $R_2\parallel C_2$, and arms 3 and 4 are $R_3$ and $R_4$. Balance means $Z_1{Y}_2=\frac{R_3}{R_4}$: $$\left(R_1+\frac{1}{j\omega C_1}\right)\left(\frac{1}{R_2}+j\omega C_2\right)=\left(\frac{R_1}{R_2}+\frac{C_2}{C_1}\right)+j\left(\omega C_2{R}_1-\frac{1}{\omega C_1{R}_2}\right)=\frac{R_3}{R_4}$$ Equating the real parts and the imaginary parts gives $$\frac{R_3}{R_4}=\frac{R_1}{R_2}+\frac{C_2}{C_1},{\omega }^2=\frac{1}{R_1{R}_2{C}_1{C}_2}$$ The imaginary-part condition fixes $\omega$, so the bridge balances at only one frequency. It is frequency-dependent, which is why it is used to measure frequency.

4. Schering bridge: arm 1 is $C_x$ in series with $r_x$, arm 2 is a lossless standard capacitor $C_2$, arm 3 is $R_3$ and arm 4 is $R_4\parallel C_4$. Then $$r_x+\frac{1}{j\omega C_x}=\frac{1}{j\omega C_2}{R}_3\left(\frac{1}{R_4}+j\omega C_4\right)=\frac{R_3{C}_4}{C_2}+\frac{R_3}{j\omega C_2{R}_4}$$ Equating the real parts gives $r_x=\frac{R_3{C}_4}{C_2}$. In the imaginary parts $\omega$ cancels, leaving $C_x=\frac{C_2{R}_4}{R_3}$. Neither balance condition contains $\omega$, so the balance is frequency-independent. Only the derived dissipation factor $D=\omega C_x{r}_x=\omega C_4{R}_4$ contains $\omega$, and that is a quantity calculated after balance, not a balance condition.

5. Wheatstone bridge: all four arms are pure resistances, so the balance condition is $R_1{R}_4=R_2{R}_3$. There is no reactance, so $\omega$ cannot appear. The balance is frequency-independent: it holds for DC and for any frequency.

6. The bridges whose balance is frequency-independent are the Maxwell, Schering and Wheatstone bridges, i.e. options (A), (C) and (D).