Average power in ac circuit and power factor - class-XII

Here, $P , = , 2 , kW , = , 2 , \times , 10^3W$
$V _{rms} , = , 233 , V, tan , \phi , = , -\dfrac{3}{4}$
$As, , P , = , \dfrac{V^2 _{rms}}{Z}$

Here, $V _0 , = , 283 , V, , R , = , 3\Omega, , L , = , 25 , \times , 10^{-3} , H$
$C , = , 400 , \mu F , = , 4 , \times , 10^{-4} F$
Maximum power is dissipated at resonance, for which 

Power is dissipated only in the Resistor , Capacitor and Inductor only store  energy. So Inductance power dissipated is Zero.

The impedance Z = sqrt(R^2 + XL^2). XL = 2 * pi * f * L = 2 * pi * (30/pi) * 0.4 = 24 ohms. Z = sqrt(10^2 + 24^2) = sqrt(100 + 576) = 26 ohms. Current I = V/Z = 6.5 / 26 = 0.25 A. Power P = I^2 * R = (0.25)^2 * 10 = 0.0625 * 10 = 0.625 W, which is 5/8 W.

Inductors and capacitors bring a phase difference between the voltage and current in the circuit, hence changing the p.f. When only a resistance is present, $Poer\ factor= 1$.
The power loss in an AC circuit$ =E _{rms} I _{rms} Power\ factor $

Maximum power ($ I^2 R )$ is obtained when $I$ is maximum ( $Z$ is minimum).

Wattless current does not contribute to the mean rate of working of the circuit.
As, power factor $= \frac{\text{true power}}{\text{apparent power}}$
                             $=cos\phi$
                             $=\frac{R}{\sqrt{R^2+(X _L-X _C)^2}}$
$\therefore$ Power factor $=cos\phi = \frac{R}{Z}$
In a non-inductive circuit, $X _L=X _C$
$\therefore$ Power factor $=cos\phi = \frac{R}{\sqrt{R^2}}=\frac{R}{R}=1$
$\therefore \phi = 0^o$
This is the maximum value of power factor. Iris a pure inductor or an ideal capacitor 
$\phi = 90^o$
$\therefore$ Power factor $= cos \phi = cos 90^o= 0$. 
Average power consumed in a pure inductor orb ideal capacitor 
$P = E _V \cdot I _V cos, 90^o = zero$.
Therefore, current through pure L or pure C; which consumes no power for its maintenance in the circuit is called ideal current or wattless current.

The power cables have some resistance. 

$P= V _{rms} \times I _{rms} \times \cos \phi$

$P=\cfrac { { V } _{ rms }^{ 2 } }{ R } cos\phi =\cfrac { { 220 }^{ 2 } }{ 200 } cos30°=209W$

$\begin{array}{l}\omega L = 2R\\frac{1}{{\omega C}} = \frac{1}{3}\left( {\omega L} \right)\\omega L - \frac{1}{{\omega C}} = 2R - \frac{{2R}}{3} = \frac{{4R}}{3}\\tan \phi  = \frac{{4R}}{3} \times \frac{1}{R} = \frac{4}{3}\\cos \phi  = \frac{1}{{\sqrt {1 + {{\tan }^2}\phi } }} = \frac{1}{{\sqrt {1 + \frac{{{4^2}}}{{{3^2}}}} }} = \frac{3}{5} = 0.6\end{array}$

Power dissipated in an AC circuit is given by P = V(rms) I(rms) cos(phi), where cos(phi) is the power factor. For a pure inductor or pure capacitor, the phase angle is 90 degrees, resulting in a power factor of zero and zero average power consumption. Only a resistor provides a non-zero power factor (cos(0) = 1) and thus maximum power consumption.

The power factor of an L-R circuit is cos(phi_old) = R / Z_old, where Z_old = sqrt(R^2 + X_L^2) = sqrt(R^2 + (3R)^2) = R sqrt(10). When a capacitor with X_C = R is added in series, the net reactance becomes X_L - X_C = 3R - R = 2R. The new impedance is Z_new = sqrt(R^2 + (2R)^2) = R sqrt(5). The new power factor is cos(phi_new) = R / Z_new = 1 / sqrt(5). The ratio of new to old power factor is (R / sqrt(5)) / (R / sqrt(10)) = sqrt(10 / 5) = sqrt(2).

In an LR circuit, Z = sqrt(R^2 + XL^2). Given XL = R, Z = sqrt(2)R. Current amplitude I0 = E0 / Z = E0 / (sqrt(2)R). Average power P = (1/2) * I0^2 * R = (1/2) * (E0^2 / 2R^2) * R = E0^2 / 4R.

In an AC circuit, the current is sinusoidal, so its average value over a complete cycle is zero. The average of the square of the current is not zero (it is I_rms^2), and average power is not necessarily zero (unless the circuit is purely reactive).

In an RL series circuit, the total voltage V is the phasor sum of the voltage across the resistor (VR) and the inductor (VL). V = sqrt(VR^2 + VL^2) = sqrt(60^2 + 80^2) = sqrt(3600 + 6400) = sqrt(10000) = 100 V.

Resonant frequency w0 = 1 / sqrt(LC) = 1 / sqrt(3 * 27 * 10^-6) = 1 / sqrt(81 * 10^-6) = 1 / (9 * 10^-3) = 1000 / 9 = 111.1 rad/s. Q-factor = (1/R) * sqrt(L/C) = (1/7.4) * sqrt(3 / (27 * 10^-6)) = (1/7.4) * sqrt(1/9 * 10^6) = (1/7.4) * (1000/3) = 1000 / 22.2 = 45.

Watt-less power in an AC circuit is basically power supposed to be generated by inductive and capacitive reactance and since they are not resistor they generate any heat , and these power wasted is called watt-less power and its phase angle is always $90^o$ as it has only capacitor and inductor.

Circuit will be capacitive if , total reactance =$ X _L - X _C < 0$

for $100V, 50 W $lamp , $ R = \frac{100^2}{50} = 200 \Omega $
capacitor put in series should be such that, 
$ V _R = \frac{ R} { \sqrt{ R^2 + X _C^2 } } V _s= 100 $
$ \Rightarrow  \frac{1}{C \omega} =  X _C= \sqrt{3} R $
$ C=\frac{1}{ \omega \sqrt{3} R } =\frac{50}{\pi \sqrt{3} } \mu F $

Power, $\displaystyle P={ I } _{ r.m.s }\times { V } _{ r.m.s }\times \cos { \phi  } $
In the given problem, the phase difference between voltage and current is p/2. Hence
$\displaystyle P={ I } _{ r.m.s }\times { V } _{ r.m.s }\times \cos { \left( { \pi  }/{ 2 } \right)  } =0\ $

$L = 20 mH$ $C = 100 \mu F$ $R = 50 \Omega$

After connecting the capacitor Irms will charge because impedance is changed.
$\therefore$ A is false

Power Factor:

In an RC circuit, the impedance Z = sqrt(R^2 + Xc^2). Given Z = 10 and R = 6, then 100 = 36 + Xc^2, so Xc = 8. The power factor is cos(phi) = R/Z = 6/10 = 0.6.

$i=100sin 100tmA$
$v=100 sin(100t+\frac {x}{3})V$
$P=I _{rms}\times V _{rms} cos\theta$
$=\frac {100}{\sqrt 2}\times \frac {100}{\sqrt 2}\times \frac {1}{2}\times 10^{-3}W$
$=\frac {10}{4}=2\cdot 5W$