A wattmeter’s coil resistances are 0.01 Ω and 1000 Ω, with negligible inductances. Connected as shown, it measures a 30 V load drawing 25 A at pf 0.8. The percentage error in its reading is ________.
The connection shown includes the voltage-coil current in the current-coil measurement, so the wattmeter reads load power plus voltage-coil dissipation.
Load power is 30×25×0.8=600 W. Voltage-coil loss is V²/Rp=30²/1000=0.9 W. Thus percentage over-reading is 100×0.9/600=**0.15%**, within 0.14–0.16.
The current-coil resistance would govern the alternative connection's error; it is not the included loss in this arrangement.
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