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Power in ac circuits - class-XII

Description: power in ac circuits
Number of Questions: 25
Created by:
Tags: physics electricity and magnetism electromagnetic induction and alternating currents alternating current
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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:

  1. 15 $\Omega$

  2. 20 $\Omega$

  3. 25 $\Omega$

  4. 30 $\Omega$


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

  1. 51.5

  2. 50.7

  3. 51.1

  4. 50.3


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

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

  2. $2LI^{2}$

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

  4. Zero


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 :

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

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

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

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


Correct Option: A

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

  1. $C$

  2. $L$

  3. $R$

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


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

  1. $8\mu F$

  2. $4\mu F$

  3. $2\mu F$

  4. $1\mu F$


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:

  1. Non-ideal current

  2. Wattless current

  3. Convectional current

  4. Inductance current


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 :

  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


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?

  1. $10^{4}$

  2. $10^{3}$

  3. $10^{2}$

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


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 :

  1. $342\ W$

  2. $305\ W$

  3. $209\ W$

  4. $242\ W$


Correct Option: C
Explanation:

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

In a series LCR circuit,the inductive reactance is twice the resistance and the capacitance reactance is ${\frac{1}{3}^{rd}}$ the inductive reactance. The power factor of the circuit is:

  1. $0.5$

  2. $0.6$

  3. $0.8$

  4. $1$


Correct Option: B
Explanation:

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

In series L.C.R circuit resonance occurs at frequency $f=f _0$, If at this moment amplitude of current is $I _0$ then calculate wattles current at the same moment:-

  1. $\sqrt 2 \ I _0$

  2. $\dfrac{I _0}{\sqrt2}$

  3. $\dfrac{I _0}{2\sqrt2}$

  4. $Zero$


Correct Option: B

Which of the following device in alternating circuit provides maximum power

  1. Only capacitor

  2. Capacitor and resistor

  3. Only inductor

  4. Only resistor


Correct Option: A

An alternative current, L.R circuit comprises of an inductor, whose reactance $X _L = 3R$, where $R$ is the resistance of the circuit. If a capacitor, whose reactance $X _C = R$ is connected in series then what will be the ratio of the new and the old power factor?

  1. $\sqrt{2}$

  2. $\dfrac{1}{\sqrt{2}}$

  3. $2$

  4. $1$


Correct Option: B

In an $LR$-circuit, the inductive reactance is equal to the resistance $R$ of the circuit. an e.m.f. $E=E _{0}\ cos(\omega t)$ applied to the circuit. The power consumed in the circuit is

  1. $\dfrac{E^{2} _{0}}{R}$

  2. $\dfrac{E^{2} _{0}}{2R}$

  3. $\dfrac{E^{2} _{0}}{4R}$

  4. $\dfrac{E^{2} _{0}}{8R}$


Correct Option: C

In general in an alternating current circuit for a complete cycle

  1. The average value of current is zero

  2. The average value of square of the current is zero

  3. Average power dissipation is zero

  4. All of these


Correct Option: A

In an a.c. circuit consisting of resistance $R$ and inductance $L$, the voltage across $R$ is $60$ volt and that across $L $ is $80$ Volt.The total Voltage across the combination is 

  1. $140 V$

  2. $20 V$

  3. $100 V$

  4. $70 V$


Correct Option: C

Find the resonant frequency and $Q-factor$ of a series $LCR$ circuit with $L = 3.0\ H, C = 27\mu F$ and $R = 7.4\Omega$.

  1. $111\ rad/s, 45$

  2. $111\ rad/s, 90$

  3. $55.5\ rad/s, 45$

  4. $55.5\ rad/s, 90$


Correct Option: A

The self inductance of a choke coil is mH. when it is connected with a 10 VDC source then the loss of power is 20 watt. When it connected with 10 volt AC source loss of power is 10 watt. The frequency of AC source will bw-

  1. 50 Hz

  2. 60 Hz

  3. 80 Hz

  4. 100 Hz


Correct Option: A
Explanation:

$\begin{array}{l} When\, \, \, dc\, \, is\, \, pass\, \, through\, \, the\, \, inductor\, \, no\, \, induv\tan  ce\, \, effect\, \, will\, \, be\, \, their\, \, only \ resis\tan  ce\, \, will\, \, be\, their\, \, resis\tan  ce\, \, of\, \, the\, \, inductor\, \, be\, \, R \ 20=\frac { { 100 } }{ R }  \ R=5\Omega  \ Xl=\omega \times 0.001 \ power\, \, =\frac { { { v^{ 2 } } } }{ z } \cos  \phi  \ Z=\sqrt { 25+{ \omega ^{ 2 } }\times { { 10 }^{ -6 } } }  \ \cos  \phi =\frac { R }{ Z } =\frac { 5 }{ Z }  \ 10=\frac { { 100 } }{ Z } \frac { 5 }{ Z } =\frac { { 500 } }{ { { Z^{ 2 } } } } =\frac { { 500 } }{ { 25 } } +{ \omega ^{ 2 } }\times { 10^{ -6 } } \ 25+{ \omega ^{ 2 } }\times { 10^{ -6 } }=50 \ { \omega ^{ 2 } }=25\times { 10^{ 6 } } \ \omega =5000 \end{array}$
$=50 Hz$
Hence, Option $A$ is correct answer.

For watt-less power in an $AC$ circuit the phase angle between the current and voltage is

  1. $0^o$

  2. $90^o$

  3. $45^o$

  4. Not possible


Correct Option: B
Explanation:

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.

For an $LCR$ series circuit with an A.C. source of angular frequency $\omega$, which statement is correct?

  1. Circuit will be capacitive if $\omega > \displaystyle\frac{1}{\sqrt{LC}}$

  2. Circuit will be capacitive if $\omega = \displaystyle\frac{1}{\sqrt{LC}}$

  3. Power factor of circuit will be unity if capacitive reactance equals inductive reactance

  4. Current will be leading voltage if $\omega > \displaystyle\frac{1}{\sqrt{LC}}$


Correct Option: C
Explanation:

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

$ \Rightarrow  X _C > X _L \Rightarrow  \frac{1}{C \omega} >  L \omega  \Rightarrow \omega < \frac{1}{ \sqrt{ L C }} $

Current will be leading the voltage if the circuit is capacitive , i.e., $ \omega   < \frac{1}{ \sqrt{ L C }} $

Power factor= $ \dfrac{ R } { Z } $   ; pf factor will be unity if $ R= Z \Rightarrow X _L=X _C $ , i.e., capacitive reactance is equal to inductive reactance.

A $50\space W$, $100\space V$ lamp is to be connected to an AC mains of $200\space V, \space 50\space Hz$. What capacitor is essential to be put in series with the lamp?

  1. $\displaystyle\frac{25}{\sqrt2}\mu F$

  2. $\displaystyle\frac{50}{\pi\sqrt3}\mu F$

  3. $\displaystyle\frac{50}{\sqrt2}\mu F$

  4. $\displaystyle\frac{100}{\pi\sqrt3}\mu F$


Correct Option: B
Explanation:

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 $

In an A.C. circuit, the current flowing in inductance is $\displaystyle I=5\sin { \left( 100t-{ \pi  }/{ 2 } \right)  } $ ampers and the potential difference is V = 200 sin (100 t) volts. The power consumption is equal to 

  1. 1000 watt

  2. 40 watt

  3. 20 watt

  4. Zero


Correct Option: D
Explanation:

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\ $

An inductor $20$ mH, a capacitor $100$ $\mu$F and a resistor $50$ $\Omega$ are connected in series across a source of emf, V$=10$ $\sin 314$t. The power loss in the circuit is?

  1. $2.74$ W

  2. $0.79$ W

  3. $1.13$ W

  4. $0.43$ W


Correct Option: B
Explanation:

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

$V = 10 sin(314 t)$
$V _0= 10$, $\omega = 314$
$X _L = wL= 314 \times 20\times 10^{-3}= 6.28 \Omega$
$X _C = \dfrac{1}{\omega C}=31.8 \Omega$
$Z = \sqrt{R^2 + (X _C- X _L)^2} = 56.1 $
$Power \ loss P=\dfrac{V _0^2 R}{2 Z^2}= 0.79 W$


Assertion: A resistance is connected to an ac source. Now a capacitor is included in the series circuit. The average power absorbed by the resistance will remain same.  

Reason: By including a capacitor or an inductor in the circuit average power across resistor does not change.

  1. A and R both are true and R is correct explanation of A

  2. A and R both are true but R is not the correct explanation of A

  3. A is true R is false

  4. A is false and R is true


Correct Option: D
Explanation:

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

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