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The resistance of two coils of a wattmeter are 0–0·1 ohm and 1000 ohms respectively and both are non-inductive. The load current is 20 A and the voltage across the load is 30 V. In one of the two ways of connecting the voltage coil, the error in the reading would be:
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- 0·1% too high
- 0·2% too high
- 0·15% too high
- zero
- 0·1% too high
Correct Option: C
107. (c) Let the resistance of the current coil is
rcc = 0·01Ω
(current coil is always low resistance as compare to voltage coil)
and the resistance of voltage coil is rvc = 1000Ω
There are two ways in which the current coil and voltage coil can be connected. The wattmeter reads high by an amount equal to power loss in current coil in one case and in voltage coil in the other. The two possible power losses are:
(1) Ic2 rcc = 202 × 0·01 = 4W
(2)
| = | ||
| rvc | 1000 |
= 0.9 W
The two percentage errors are
(i) 4/(20 × 30) × 100 = ·67%
(ii) 0·9/(20 × 30) × 100 = 0·15%
