GATE 2018 IN Q.48: solution and answer

GATE 2018 · INQ.48 · 2 marks · NAT

A high Q coil having distributed (self) capacitance is tested with a Q-meter. First resonance at $ω_{1} = 1 0^{6}$ rad/s is obtained with a capacitance of 990 pF. The second resonance at $ω_{2} = 2 \times 1 0^{6}$ rad/s is obtained with a 240 pF capacitance. The value of the inductance (in mH) of the coil is (up to one decimal place) ______.

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Answer and solution

ANSWER

The inductance of the coil is $L = 1.0 mH$ , with a self-capacitance of $C_{d} = 10 pF$ .

Official answer: 0.9 to 1.1Source: Official answer key, GATE 2018 IN (IIT Guwahati)

In a Q-meter the coil's distributed (self) capacitance $C_{d}$ is in parallel with the tuning capacitor $C$ . Resonance therefore occurs when $ω^{2} L (C + C_{d}) = 1$ .
First resonance, at $ω_{1} = 1 0^{6} rad/s$ with $C_{1} = 990 pF$ : $ω_{1}^{2} L (C_{1} + C_{d}) = 1$
Second resonance, at $ω_{2} = 2 \times 1 0^{6} rad/s$ with $C_{2} = 240 pF$ : $ω_{2}^{2} L (C_{2} + C_{d}) = 1$
Both products equal 1, so set them equal. $L$ cancels, and $ω_{2}^{2} = 4 ω_{1}^{2}$ : $C_{1} + C_{d} = 4 (C_{2} + C_{d}) \Rightarrow 990 + C_{d} = 960 + 4 C_{d} \Rightarrow C_{d} = 10 pF$ This is the standard doubled-frequency result $C_{d} = \frac{C _{1} - 4 C _{2}}{3} = \frac{990 - 960}{3} = 10 pF$ .
Total resonating capacitance at the first resonance: $C_{1} + C_{d} = 1000 pF = 1000 \times 1 0^{- 12} F = 1 0^{- 9} F$ .
Solve the first resonance condition for $L$ : $L = \frac{1}{ω _{1}^{2} ( C _{1} + C _{d} )} = \frac{1}{( 1 0 ^{6} ) ^{2} \times 1 0 ^{- 9}} = \frac{1}{1 0 ^{12} \times 1 0 ^{- 9}} = \frac{1}{1 0 ^{3}} = 1 0^{- 3} H$
Check with the second resonance: $C_{2} + C_{d} = 250 pF = 2.5 \times 1 0^{- 10} F$ , so $ω_{2}^{2} (C_{2} + C_{d}) = 4 \times 1 0^{12} \times 2.5 \times 1 0^{- 10} = 1 0^{3} \Rightarrow L = 1 0^{- 3} H$ Both resonances give the same inductance.
Convert to the unit asked for: $L = 1 0^{- 3} H = 1.0 mH$ .

KEY IDEA

A coil's self-capacitance $C_{d}$ adds to the Q-meter's tuning capacitance, so $ω^{2} L (C + C_{d}) = 1$ at every resonance. Two resonances give two equations. Equating them removes $L$ and gives $C_{d}$ . For a frequency ratio of 2, $C_{d} = (C_{1} - 4 C_{2}) /3$ . Put $C_{d}$ back into either equation to get $L$ .

WATCH OUT FOR

The usual slip is the powers of ten in the last step. $1000 pF$ is $1 0^{- 9} F$ , not $1 F$ , so $ω_{1}^{2} (C_{1} + C_{d}) = 1 0^{12} \times 1 0^{- 9} = 1 0^{3}$ . That gives $L = 1 mH$ , not $1 H$ (1000 mH). Two other errors: leaving out $C_{d}$ altogether, which gives $L = 1/ (1 0^{12} \times 990 \times 1 0^{- 12}) \approx 1.01 mH$ from the first resonance and a different value from the second, and subtracting $C_{d}$ from $C$ when it should be added.

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