A 4.0 ampere ammeter has a resistance of 0.01 Ω. Determine the efficiency of the instrument.

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UPPCL JE Electrical 7 Sept 2021 Official Paper (Shift 2)
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1. 25 A/W
2. 25.01 A/W
3. 16 A/W
4. 1.6 A/W

Option 1 : 25 A/W

Detailed Solution

Effiiciency of Instruments:

The efficiency of instruments is the ratio of voltage and current rating to the power rating.

For ammeter, the current rating is considered and for voltmeter, the voltage rating is considered.

Formula:

For Ammeter: P = I2R

For Voltmeter: $$P=\frac{V^2}{R}$$

Application:

The given instrument is ammeter,

I = 4 A

R = 0.01 Ω

Hence, P = I2R = 42 × 0.01 = 0.16 W

Hence, efficiency = I/P = 4/0.16 = 25 A/W