Measurement Of Power
where, = RMS value of voltage
= RMS value of current
and cosø = Power factor of the load
Elctrodynamometer Type Wattmeter :
And the other coils is “Moving coils” or “Pressure coil” or “Voltage coil” or “Pontential coil (PC)“.
Let rp and Lp are the resistance and inductance of the pressure coil or moving coil respectively and Rp = (rp + R) = total resistance of PC circuit
l = current in CC circuit = lCC
= impedance of PC circuit.
Let load p.f. cosø lagging i.e. lags behind V by an angle ø .
and controlling torque = Tc = K θ
At null deflection, Tc = Td
since, θ ∝ Vl cos ø i.e.
θ ∝ Power (P)
Measurement of 3-Phase Power with two wattmeter method :
P1 = i1(v1 – v3)
Instantaneous reading of W2 wattmeter
P2 = i2(v2 – v3)
Sum of instantaneous reading of two wattmeter is, P = P1 + P2
P = i1(v1 – v3) + i2(v2 – v3)
From KCL at node ‘O’, i1 + i2 + i3 = O, or (i1 + i2) = – i3
∴ sum of instantaneous readings of two wattmeters P = v1i1 + v2i2 + v3i3
Let V1, V2, V3, be the rms values of phase voltage and I1, I2, I3 be the rms values of phase currents.
Since the load is balanced, therefore,
Phase voltages, V1 = V2 = V3 = V(say)
Phase currents, I1 = I2 = I3 = I(say)
and, Power factor = cos ø
and reading wattmeter W2 is
Sum of reading of two wattmeter is, P = P1 + P2
P = 3 VI cosø = 3 - ø power
This shows the total power consumed by the load.
From this equation (4.38) we have also find pf = cosø.