Q21GATE 2004MCQ

A galvanometer with a full scale current of 10 mA has a resistance of 1000 $\Omega$ . The multiplying power (the ratio of measured current to galvanometer current) of 100 $\Omega$ shunt with this galvanometer is

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📖 Explanation

Full scale current of galvanometer $I_m=10mA$ Resistance of meter $R_m=1000\Omega$ Resistance of shunt $R_{sh}=100\Omega$ $I_m=\frac{IR_{sh}}{R_{sh}+R_m}$ Multiplying power

\begin{aligned} &=\frac{I}{I_m}=\frac{R_{sh}+R_m}{R_m} \\ &= \frac{100+1000}{100}=11 \end{aligned}

Q23GATE 2004MCQ

A moving coil of a meter has 100 turns, and a length and depth of 10 mm and 20 mm respectively. It is positioned in a uniform radial flux density of 200 mT. The coil carries a current of 50 mA. The torque on the coil is

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📖 Explanation

T=Torque on the coil=NBAI where,

\begin{aligned} N &= \text{Number of turns}=100 \\ B&= \text{Flux density}\\ &= 200mT =200 \times 10^{-3}T\\ A &= \text{Area of the coil} \\ &= \text{length} \times \text{depth}\\ &= (10 \times 10^{-3}) \times (20 \times 10^{-3})\\ &= 200 \times 10^{-6} m^2\\ I&= \text{Current through the coil}\\ &= 50mA=50 \times 10^{-3}\\ T&= NBAI \\ &=100 \times (200 \times10^{-3}) \times (200 \times 10^{-6}) \\ &= \times (50 \times 10^{-3})\\ &=2 \times 10^{-4} N-m=200\mu N-m \end{aligned}

Q24GATE 2003MCQ

The effect of stray magnetic field on the actuating torque of a portable instrument is maximum when the operating field of the instrument and the stray fields are

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Q25GATE 2003MCQ

A Manganin swap resistance is connected in series with a moving coil ammeter consisting of a milli-ammeter and a suitable shunt in order to

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📖 Explanation

Coil is made of copper A swamping resistance ( $R_{sw}$ ) of manganin (which has a negligible temperature coefficient) having a resistance of 20 to 30 times the coil resistance is connected in series with the coil and a shunt of manganin is connected across this combination. Since copper forms a samll fraction of the series combination, the proportion in which the currents would divide between the meter and the shunt would not change appreciably with the change in temperature.

Q26GATE 2003MCQ

The inductance of a certain moving-iron ammeter is expressed as L = $10+30 -\frac{\theta ^{2}}{4} \; \mu H$ , where $\theta$ is the deflection in radians from the zero position. The control spring torque is $25 \times 10^{-6}$ Nm/radian. The deflection of the pointer in radian when the meter carries a current of 5 A, is

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📖 Explanation

Deflection torque in moving-iron ammeter, $T_d=\frac{1}{2}I^2\frac{dL}{d\theta }$ Inductance, $L=\left ( 10+3\theta -\frac{\theta ^2}{4} \right )\mu H$ Rate of change of inductance with deflection $\frac{dL}{d\theta }=\frac{d}{d\theta }\left ( 10+3\theta -\frac{\theta ^2}{4} \right )$ $\;\;\;=\left ( 3- \frac{\theta }{2}\right )\mu H/rad$ Current , $I=5A$ Deflecting torque,

\begin{aligned} T_d &=\frac{1}{2}I^2\frac{dL}{d\theta } \\ T_d &= \frac{1}{2} \times 5^2 \times \left ( 3-\frac{\theta }{2} \right ) \times 10^{-6}\\ &= \frac{25}{2}\left ( 3-\frac{\theta }{2} \right )\times 10^{-6} Nm \end{aligned}

Controlling torque $T_c=k\theta =25 \times 10^{-6}\theta$ At equilibrium,

\begin{aligned} T_c &= T_d\\ 25 \times 10^{-6}\theta &= \frac{25}{2}\left ( 3-\frac{\theta }{2} \right )\times 10^{-6} \\ \frac{5\theta }{2} &= 3\\ \theta &= 1.2 \text{rad} \end{aligned}

Q27GATE 2003MCQ

A rectifier type ac voltmeter of a series resistance $R_{s}$ , an ideal full-wave rectifier bridge and a PMMC instrument as shown in figure. The internal. resistance of the instrument is 100 $\Omega$ and a full scale deflection is produced by a dc current of 1 mA. The value of $R_{s}$ required to obtain full scale deflection with an ac voltage of 100 V (rms) applied to the input terminals is

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📖 Explanation

$I_{FS}=$ Current required to produce full scale deflection d.c. sensitivity is, $S_{dc}=\frac{1}{I_{(FS)}} =\frac{1}{1 mA}=10^3\Omega /V$ For full wave rectifier a.c. sensitivity $S_{ac}=0.9S_{dc}=900\Omega /V$ Resistance of multiplier, $R_S=s_{ac}V-R_m-2R_d$ Since, diodes are ideal, $R_d=0$ then, $R_S=900 \times 100-100 =89.9k\Omega$

Q28GATE 2001MCQ

A $100\mu A$ ammeter has an internal resistance of 100 $\Omega$ . For extending its range to measure $500\mu A$ , the shunt resistance required is of (in $\Omega$ )

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📖 Explanation

$R_{sh}=\frac{R_a}{m-1}$ $\;\;\;=\frac{100}{\frac{500}{100}-1}=25.0\Omega$