GATE 2024 IN Q.56: solution and answer

GATE 2024 · INQ.56 · 2 marks · NAT

A shielded cable with $C_{stray} = 20 pF$ and $R_{wire} = 10 Ω$ is used to connect the inductive sensors as shown in the figure. The RMS value of $V_{out}$ is ____ V (rounded off to two decimal places).

Note: Assume all components are ideal, and sensors are not magnetically coupled.

GATE 2024 IN question 56: Two variable inductors (inductive sensors, each drawn with a diagonal arrow) are in series. — Figure from the GATE 2024 IN paper. It is part of the question.

Answer and solution

ANSWER

$V_{out} = 4 sin (2000 π t) V$ , so $V_{out,rms} = 22 \approx 2.83 V$ .

Official answer: 2.81 to 2.85Source: Official answer key, GATE 2024 IN (IISc Bengaluru)

What the guard does. The unity-gain buffer copies the voltage of the cable's inner wire onto the shield, so both plates of $C_{stray}$ sit at the same potential. With zero voltage across it, $C_{stray}$ carries no current and has no effect on the signal. This is a driven shield. (Even without the guard its effect would be negligible here. At $f = 1 kHz$ , $X_{C} = \frac{1}{2 π ( 1000 ) ( 20 \times 1 0 ^{- 12} )} \approx 7.96 M Ω$ , while the sensor divider's Thevenin impedance is only $ω (L_{1} ∥ L_{2}) \approx 8.4 Ω$ .)
The cable draws no current. An ideal op-amp input and an ideal buffer input draw zero current, and $C_{stray}$ carries none. So no current flows in $R_{wire}$ or in the $2 Ω$ input resistor. Both drop $0 V$ , and the + input sits at the sensor junction voltage $V_{m}$ .
The sensors form an unloaded inductive divider. $L_{1}$ and $L_{2}$ carry the same current and are not magnetically coupled, so each has impedance $j ω L$ with $ω = 2000 π rad/s$ ( $f = 1 kHz$ ). The two sources are in phase (one is the negative of the other), so the $j ω$ factors cancel and the divider acts on the amplitudes directly.
Superposition (Millman's theorem) at the junction: $V_{m} = \frac{V _{s 1} ( j ω L _{2} ) + V _{s 2} ( j ω L _{1} )}{j ω ( L _{1} + L _{2} )} = \frac{6 ( 4 ) + ( - 6 ) ( 2 )}{2 + 4} sin (2000 π t) = 2 sin (2000 π t) V$
Check with KVL: $I j ω (L_{1} + L_{2}) = V_{s 1} - V_{s 2} = 12 sin (2000 π t)$ . The drop across $L_{1}$ has amplitude $12 \cdot \frac{2}{6} = 4$ , so $V_{m} = 6 - 4 = 2$ (amplitude). This agrees with the previous step.
Non-inverting amplifier gain. Assume an ideal op-amp (virtual short between the inputs, no input current) and an output that stays inside the supply rails: $A_{v} = 1 + \frac{R _{f}}{R _{g}} = 1 + \frac{2 Ω}{2 Ω} = 2$
Output: $V_{out} = 2 \times 2 sin (2000 π t) = 4 sin (2000 π t) V$
RMS value: $V_{out,rms} = \frac{4}{2} = 22 \approx 2.83 V$

KEY IDEA

The buffer holds the shield at the inner wire's voltage, so $C_{stray}$ has no voltage across it and carries no current. The ideal amplifier inputs draw no current either, so $R_{wire}$ and the $2 Ω$ input resistor drop nothing. The op-amp therefore sees the unloaded inductive-divider voltage $V_{m} = \frac{V _{s 1} L _{2} + V _{s 2} L _{1}}{L _{1} + L _{2}} = 2 sin (2000 π t)$ . A non-inverting gain of 2 makes the peak 4 V, and $4/ 2 \approx 2.83 V$ RMS.

WATCH OUT FOR

Swapping the divider weights. The junction voltage weights $V_{s 1}$ by the FAR inductor $L_{2}$ , not by $L_{1}$ . Writing $\frac{6 ( 2 ) + ( - 6 ) ( 4 )}{6}$ gives $- 2$ , which has the same magnitude only by coincidence, so check the sign with KVL. Other slips: using an inverting gain of $- R_{f} / R_{g} = - 1$ instead of the non-inverting $1 + R_{f} / R_{g} = 2$ (answer $1.41$ ); loading the divider with $R_{wire}$ , the $2 Ω$ resistor or $C_{stray}$ even though no current flows; and reporting the peak (4 V) instead of the RMS value.

From the GATE 2024 IN paper. Every question of this paper · Download the full paper · All 2 Noise & shielding (IN) questions, with notes