The current taken from a 230 v, 50 hz supply is measured as 10 a with a lagging pf of 0.7
a. Capacitor is connected in parallel with the load. The true power
Answers
Despite all this the product of current and voltage is still used in AC circuits and is called apparent power (VA), giving a value of voltamperes (VA). The term apparent power is misleading because it suggests that the apparent power is dissipated, however as we found previously the power dissipated in an AC circuit, called active, true or real power (in the units of watts), is given by:
\mathbf{P=IU\boldsymbol{\cos}\phi=watts\; (W)}
The apparent power is given by:
\mathbf{VA=UI=voltamperes\; (VA)}
These definitions are true under all circumstances and if the supply is sinusoidal:
\text{Power Factor }=\dfrac{\text{active power}}{\text{apparent power}}=\dfrac {UI\cos \phi}{\text{apparent power}}=\cos \phi
From section 10.1 we can add that:
\mathbf{PF=\dfrac{P}{UI}=\boldsymbol{\cos}\phi=\dfrac{U_R}{Z}}
In a predominantly inductive series circuit, where current lags voltage, the power factor is called a lagging power factor. Similarly, in a predominately capacitive series circuit, where current leads voltage, the power factor is called a leading power factor. The power factor can vary between definite limits, being 1 (unity) for purely resistive circuits, where the phase angle is 0° and P = UI; or 0 for purely reactive (inductive or capacitance) circuits, where the phase angle is 90° and P = 0. Note:
if PF = 1 (i.e. purely resistive circuit) active power = apparent power = UI
if PF = 0 (i.e. purely inductive or capacitive circuit) active power = reactive power = UI (section 11.3)
Example
An AC single-phase motor takes 5A at 0.7 lagging power factor when connected to a 240V, 50Hz supply. Calculate the power input to the motor. If the motor efficiency is 70%, calculate the output.
P=UI\cos \phi=240\times 5\times 0.7=0.48W
\begin{array}{lll} \text{Output Power } & = & \text{ input power}\times \text{ efficiency}\\ & = & 840\times 70/100=588W \end{array}
Example
A 200V AC circuit comprises a 40Ω resistor in series with a capacitor of reactance 30Ω. Calculate the current and the power factor.
Z=\surd \left (R^2+X_C^2 \right )=50\Omega
I=\dfrac {U}{Z}=\dfrac {200}{50}=4A
PF=\dfrac {R}{Z}=\dfrac {40}{50}=0.8\quad \text{leading}
Alternative, power factor could have been calculated from values of true and apparent power.
VI=200\times 4=800VA
P=UI\cos \phi =UI\dfrac {R}{Z}=800\times \dfrac {40}{50}=640W
PF=\dfrac {P}{UI}=\dfrac {640}{800}=0.8\quad \text{leading}