The power factor of an AC electrical power system is defined as the ratio of the real power flowing to the load to the apparent power in the circuit. Power factor meter will measures power factor. In power factor meter there will be no controlling torque.
The meter constant of a single phase 240 V induction watt hour meter is 400 revolutions per kWh. The speed of the meter disc for a electric current of 10 amperes of 0.8 p.f. lagging will be
n = K × Power × Time in hour = 400 × 240 × 10 × 0.8 / 100 × 1 / 3600 = 12.8 rpm
The effect of stray magnetic fields on the actuating torque of a portable instrument is maximum when the operating field of the instrument and the stray fields are
Due to stray magnetic field, torque is also produced which can affect the torque produce due to operating field. If both stray magnetic field and operating field are parallel, torque due to both field become a additive.
In a two watt meter method the reading of W1 = 3 KW and W2 = 2 KW. But W2 reading was taken after reversing the electric current soil of the wattmeter. The net power in the circuit is ________________.
Since the electric current coil of W2is reversed the reading of it should be taken as - 2 KW.
Sum of two watt meters reading, i.e., 1 KW is the total power drawn by the circuit.
Measurement of low resistance methods
All these 3 methods are used for measurement of low resistance.
Two wattmeter reading are 1500 W and 700 W, what is there sum?
Sum of 2 wattmeter reading is 700 + 1500 watt = 2200 watt. Since sum of 2 readings = total power consumed by load irrespective of whether load is balanced / not.
An ammeter has reading range of 0 - 5 A and internal resistance of 0.2 ohm, in order to make the range 0 - 25 A we need to add a resistance in
Resistance have to be added in parallel to the internal resistance of ammeter.
A 1000 Ω / V, meter is used to a measure a resistance on 150 V scale. The meter resistance is
Resistance = sensitivity in ohm × voltage in volts = 1000 / 150 = 6.67 Ω.
One single phase watt meter operating on 230 V and 5 A for 5 hours makes 1940 revolutions. Meter constant in revolution is 400.The Power factor of the load will be
An average response rectifier type electric AC voltmeter has dc voltage of 10 Volt applied to it. The meter reading will be
A rectifier type instrument is calibrated to read rms values for sinusoidal, waveform Thus it shows 1.11 time and average value. Therefore reading should be = 10 × 1.11 = 11.1 Volt
The meter constant of a 240 V induction watt hour meter is 400 revolution per KWh. The speed of the meter disc for a electric current of 10 Ampere of 0.8 p.f lagging will be-
n = K × Power × Time in hour = ( 400 × 240 × 10 × 0.8) / 100 ×( 1 / 3600) = 12.8 rpm
A 0 – 15 V voltmeter has a resistance of 1000 ohms if it is desired to expand its rang to 0 – 150 V a resistance of ____________ is connected in series with it.
To improve the range Rs a high resistance is connected in series with the meter. I = 15 / 1000 A;
⇒ 150 V = 15 / 1000 × ( Rs + 1000 )
⇒ Rs = 9 kΩ
At pf = 1 1st wattmeter reads x watt and 2nd wattmeter reads y watts then
At unity pf, readings of 2 wattmeter are equal each reads half the total power. Therefore x = y
The pressure coil of an energy meter is
'Pressure coil' is an archaic term for 'voltage coil'. Pressure coils is having large no.of turns, which is connected in parallel with the supply. Large no.of turns offers high impedance i.e., highly inductive in nature(because of more no. of turns.
The pressure coil of a dynamometer type wattmeter is
Purely resistive coil is desired but it is difficult to have purely resistive pressure coil. The pressure coil has a small value of inductor due to which error occurs in wattmeter readings . That must be highly resistive.