One single phase wattmeter operating on 230V and 5A for 5 hours makes 1940 revolutions. Meter constant in revolutions is 400. What is the power factor of the land?
A. 1
B. 0.84
C. 0.73
D. 0.65
Answer: Option B
A. 1
B. 0.84
C. 0.73
D. 0.65
Answer: Option B
The spring material used in a spring control device should have the following property.
A. Should be nonmagnetic
B. Most be of low temperature coefficient
C. Should have low specific resistance
D. All of the above
Why is a MISC meter not recommended for DC measurement?
A. The meter is calibrated for AC and it’s error for DC would be high
B. The meter does not respond to DC signals
C. The error is high due to hysteresis effect
D. The error is high due to eddy current effect
A moving iron ammeter may be compensated for frequency errors by
A. A shunt resistance
B. A series inductance
C. Shunt capacitance
D. Series resistance
@Smiths saji thank for your explanation
kWh consumed=230 x 5 x 5 =5.750 kWh
meter constant= 400 rev/kWh
rev for 5 hours = 1940
kWh= rev/ meter constant= rev/rev/kWh =1940/400=4.85 kWh
Actual power factor=4.85/5.75=0.843
Energy consumed = 230 × 5 × 5 = 5750 Wh
= 5.57 kWh Revolution constant = 400 rev/kWh
∴ kWh =
1940
= 4.8
400
Power factor =
4.8
= 0.8
Aparrent power (kvah)= 230×5x5/1000=5.750
Active power (KWh)= 1940/400=4.85
P.f (Cos €) =KWh/Kvah= 4.85/5.75=0.843
kWh consumed=230 x 5 x 5 =5.750 kWh
meter constant= 400 rev/kWh
rev for 5 hours = 1940
kWh= rev/ meter constant= rev/rev/kWh =1940/400=4.85 kWh
Actual power factor=4.85/5.75=0.843
GIVEN V=230V, I=5A& T=5 Hours
meter constant =400, revolution=1940
energy (E)=REVOLUTION/ METER CONSTANT
energy (E)=1940/400 =4.85 KWH = 4850WH
Energy consumed(E) = 230v *5A * 5 hours = 5.750 kwh = 5750 WH
POWER FACTOR = 4850/5750 =0.843
Energy consumed = 230v *5A * 5 hours = 5.750 kwh
Revolution constant =400/kwh so =1940/400= 4.85
P.F = 4.85/5.750= 0.84
@hari
please explain this question sir.