Tag: average power in ac circuit and power factor

Questions Related to average power in ac circuit and power factor

An electrical device draws 2 kW power from ac mains voltage 223 V(rms). The current differs lags in phase by $\phi = tan^{-1} \left ( -\frac{3}{4} \right )$ as compared to voltage. The resistance R in the circuit is:

physics alternating current power in ac circuits average power in ac circuit and power factor power in ac circuit

  1. 15 $\Omega$

  2. 20 $\Omega$

  3. 25 $\Omega$

  4. 30 $\Omega$

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Correct Option: B
Explanation:

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

$\Rightarrow \, Z \, = \, \dfrac{V^2 _{rms}}{P} \, = \, \dfrac{(223)^2}{2000} \, =  \dfrac{49729}{2000} \, = \, 24.86 \, \Omega \, or \, Z \, = \, 25 \, \Omega$

$\tan \, \phi \, = \, \dfrac{X _C \, - \, X _L}{R} \, = \, - \dfrac{3}{4} \, \therefore \, X _C \, - \, X _L \, = \, -\dfrac{3}{4} R.$

AS, $Z^2 \, = \, R^2 \, + \, (X _C \, - \, X _L)^2$
$\therefore \, (25)^2 \, = \, R^2 \, + \, \left(-\dfrac{3}{4} \, R \right)^2$
$625 \, = \, \dfrac{25 \, R^2}{16}.$
$R^2 \, = \, \dfrac{625 \, \times \, 16}{25} \,  \, \Rightarrow \, R \, = \, 20 \, \Omega$

A voltage of peak value 283 V and varying frequency is applied to series LCR combination in which R = 3$\Omega$, L = 25 mH and C = 400$\mu$F. Then the frequency (in Hz) of the source at which maximum power is dissipated in the above is

physics alternating current power in ac circuits average power in ac circuit and power factor power in ac circuit

  1. 51.5

  2. 50.7

  3. 51.1

  4. 50.3

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Correct Option: D
Explanation:

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 

$\nu \, = \, \dfrac{1}{2\pi \sqrt{LC}} \, = \, \dfrac{1 \, \times \, 7}{2 \, \times \, 22 \, \sqrt{25 \, \times \, 10^{-3} \, \times \, 4 \, \times \, 10^{-4}}}$

$= \, \dfrac{7 \, \times \, 10^3}{44\sqrt{10}} \, = \, 50.3 \, Hz$

Power dissipated in pure inductance will be

physics alternating current power in ac circuits average power in ac circuit and power factor power in ac circuit

  1. $\dfrac {LI^{2}}{2}$

  2. $2LI^{2}$

  3. $\dfrac {LI^{2}}{4}$

  4. Zero

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Correct Option: D
Explanation:

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

A coil has a resistance $ 10 \Omega $ and an inductance of 0.4 henry. It is connected to an AC source of $ 6.5 V , \frac {30} { \pi } Hz. $ The average power consumed in the circuit, is :

physics alternating current power in ac circuits average power in ac circuit and power factor power in ac circuit

  1. $ \cfrac {5} { 8} W $

  2. $ \cfrac {4} {3} W $

  3. $ \cfrac {3} {8} W $

  4. $ \cfrac {6} {7} W $

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Correct Option: A

The power loss in an $AC$ circuit is $E _{rms}$ $I _{rms}$, when in the circuit there is only

physics alternating current power in ac circuits average power in ac circuit and power factor power in ac circuit

  1. $C$

  2. $L$

  3. $R$

  4. $L,\ C$ and $R$

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Correct Option: C
Explanation:

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 $

The self inductance of the motor of an electric fan is 10 H. In order to impart maximum powr of 50 Hz, it should be connected to a capacitance of

physics alternating current power in ac circuits average power in ac circuit and power factor power in ac circuit

  1. $8\mu F$

  2. $4\mu F$

  3. $2\mu F$

  4. $1\mu F$

Show answer
Correct Option: D
Explanation:

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

For $Z$ minimum, $X _L=X _C$, which yields
$C=\dfrac {1}{(2\pi n)^2L}=\dfrac {1}{4\pi^2\times 50\times 50\times 10}$

$\therefore C=0.1\times 10^{-5}F=1\ \mu F$

The current which does not contribute to the power consumed in an AC circuit is called:

physics alternating current power in ac circuits average power in ac circuit and power factor power in ac circuit

  1. Non-ideal current

  2. Wattless current

  3. Convectional current

  4. Inductance current

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Correct Option: B
Explanation:

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 loss is less in transmission lines, when :

physics alternating current power in ac circuits average power in ac circuit and power factor power in ac circuit

  1. Voltage is less but current is more

  2. Both voltage and current are more

  3. Voltage is more but current is less

  4. Both voltage and current are less

Show answer
Correct Option: C
Explanation:

The power cables have some resistance. 

Power lost in the wires can be calculated as $P=I^2R$ with $R$ as the resistance of the wires and $I$ as the current that passes through them.
Power at the load is $P=VI$. 
From this one can see that if  voltage is increased by say $n$ times, then only $\dfrac{1}{n}$ the current is required to deliver the same power. However, if $\dfrac{1}{n}$ current is passed on the same wires, only $\dfrac{1}{n^2}$ of the power will be lost.

If $V=100 \sin 100t$ volt, and $I=100 \sin(100t+\dfrac {\pi}{6})A$. then find the watt less power in watt?

physics alternating current power in ac circuits average power in ac circuit and power factor power in ac circuit

  1. $10^{4}$

  2. $10^{3}$

  3. $10^{2}$

  4. $2.5 \times 10^{3}{\sqrt{3}}$

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Correct Option: D
Explanation:

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


$\quad= \large\frac{V _0I _0}{\sqrt{2}\times \sqrt{2}}\times \cos \dfrac{\pi}{6}$

$\quad= \large\frac{100 \times 100}{2} \times \frac{\sqrt{3}}{2}=2.5\times 10^3\sqrt{3}W$

In a series $LCR$ circuit $K=200\ \Omega$ and the voltage and frequency of the main supply are $220\ V$ and $50\ Hz$ respectively. On taking out the capacitor from the circuit, the current leads the voltage by ${30}^{o}$. On taking out the indicator from the circuit the current leads the voltage by ${30}^{o}$. The power dissipated in the $LCR$ circuit is :

physics alternating current power in ac circuits average power in ac circuit and power factor power in ac circuit

  1. $342\ W$

  2. $305\ W$

  3. $209\ W$

  4. $242\ W$

Show answer
Correct Option: C
Explanation:

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