GATE 2018 EE Q.27: solution and answer

GATE 2018 · EEQ.27 · 2 marks · MCQ

A 0-1 Ampere moving iron ammeter has an internal resistance of 50 m $Ω$ and inductance of 0.1 mH. A shunt coil is connected to extend its range to 0-10 Ampere for all operating frequencies. The time constant in milliseconds and resistance in m $Ω$ of the shunt coil respectively are

(A)
2, 5.55
(B)
2, 1
(C)
2.18, 0.55
(D)
11.1, 2

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Answer and solution

ANSWER

Option (A): the shunt coil has a time constant of 2 ms and a resistance of $50/9 \approx 5.55 m Ω$ .

Official answer: ASource: Official answer key, GATE 2018 EE (IIT Guwahati)

Time constant of the meter coil: $τ_{m} = \frac{L _{m}}{R _{m}} = \frac{0.1 \times 1 0 ^{- 3} H}{50 \times 1 0 ^{- 3} Ω} = 2 \times 1 0^{- 3} s = 2 ms$ .
The meter and the shunt are in parallel, so the current divides inversely as the branch impedances: $\frac{I _{m}}{I _{s h}} = \frac{R _{s h} + j ω L _{s h}}{R _{m} + j ω L _{m}} = \frac{R _{s h}}{R _{m}} \cdot \frac{1 + j ω τ _{s h}}{1 + j ω τ _{m}}$ For this ratio to be the same at every frequency, the factor $\frac{1 + j ω τ _{s h}}{1 + j ω τ _{m}}$ must equal 1 for all $ω$ , so the two time constants must be equal: $\frac{L _{s h}}{R _{s h}} = \frac{L _{m}}{R _{m}}$ .
So the shunt coil's time constant is $τ_{s h} = τ_{m} = 2 ms$ .
With equal time constants, $\frac{I _{m}}{I _{s h}} = \frac{R _{s h}}{R _{m}}$ at every frequency, so the ordinary DC shunt formula applies. The multiplying power is $m = \frac{I}{I _{m}} = \frac{10}{1} = 10$ , and $I = I_{m} + I_{s h} = I_{m} (1 + \frac{R _{m}}{R _{s h}})$ gives $R_{s h} = \frac{R _{m}}{m - 1}$ .
$R_{s h} = \frac{50 m Ω}{10 - 1} = \frac{50}{9} m Ω \approx 5.56 m Ω$ (option (A) prints this as 5.55, cutting off rather than rounding the last digit). The shunt inductance follows as $L_{s h} = τ_{s h} R_{s h} = 2 \times 1 0^{- 3} \times \frac{50}{9} \times 1 0^{- 3} \approx 0.0111 mH$ .
The pair (2 ms, 5.55 m $Ω$ ) is option (A).

KEY IDEA

For a shunt to divide the current the same way at every frequency, the shunt branch must have the same $L / R$ time constant as the meter coil. The impedance ratio is then a pure resistance ratio, so the DC formula $R_{s h} = R_{m} / (m - 1)$ still holds.

WATCH OUT FOR

One common slip is to use $R_{s h} = R_{m} / m = 5 m Ω$ instead of $R_{m} / (m - 1)$ . The meter carries 1 A of the 10 A, so the shunt carries only 9 A. Another is to think frequency independence means the shunt must be purely resistive. A non-inductive shunt makes the reading depend on frequency, because the meter's reactance $ω L_{m}$ grows with $ω$ . The shunt needs the same time constant as the meter, not zero inductance.

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GATE 2018 EE Q.27: solution and answer

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