Tag: power in ac circuits

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

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 $\omega > \displaystyle\frac{1}{\sqrt{LC}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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. $\displaystyle\frac{100}{\pi\sqrt3}\mu F$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $

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

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
Reveal answer Fill a bubble to check yourself
D Correct answer
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\ $

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

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $
$Power \ loss P=\dfrac{V _0^2 R}{2 Z^2}= 0.79 W$


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

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
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

For an LCR circuit, the power transferred from the driving source to the driven oscillator is $P = I^2Z cos\phi$ Then

  1. the power factor $cos \phi &gt; 0, P &gt; 0$.

  2. the driving force can give no energy to the oscillator (P = 0) in some cases.

  3. the driving force cannot syphon Out (P &lt; 0) the energy out of oscillator.

  4. all of these.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Power Factor:

(1) It may be defined as cosine of the angle of lag or lead $(i.e, \cos \phi)$
(2) It is also defined as the ratio of resistance and impedance $\left(i.e, \dfrac{R}{Z}\right)$

(3) The ratio $= \dfrac{\text{True power}}{\text{Apparnet power}} = \dfrac{W}{VA} = \dfrac{kW}{kVA} = \cos \phi$

In the given problem power transferred,
$P = I^2Z \cos \phi$
where $I$ is the current, $Z=$ Impedance and $\cos \phi$ is power factor

(a) As power factor, $\cos \phi = \dfrac{R}{Z}$
where $R > 0$ and $Z > 0$
$\Rightarrow \cos \phi > 0 \Rightarrow P > 0$

(b) When $\phi = \dfrac{\pi}{2}$  (in case of $L$ of $C$), $P = 0$.

(c) From (a), it is clear that $P < 0$ is not possible 

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

A resistance $R\Omega$ is connected in series with capacitance $C$ Farad value of impedance of the circuit is $10\Omega$ and $R=6\Omega$ so, find the power factor of circuit.

  1. $0.4$

  2. $0.6$

  3. $0.67$

  4. $0.9$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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.

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

In an AC circuit $V$ and $I$ are given by $V=100\sin{\left(100t\right)}$volt, $I=100\sin{\left(100t+\dfrac{\pi}{3}\right)}$amp the power dissipated in the circuit is

  1. $5.0\ kW$

  2. $2.5\ kW$

  3. $1.25\ kW$

  4. $zero$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$P={V} _{rms}\times{I} _{rms}\times\cos{\phi}$
$=\dfrac{{V} _{\circ}{I} _{\circ}}{2}\cos{\phi}$
$=\dfrac{100\times 100}{2}\times \dfrac{1}{2}$
$=2500$W
$=2.5$kW
Multiple choice physics alternating current power in ac circuits average power in ac circuit and power factor power in ac circuit

The current and voltage functions in an AC circuit are$i = 100\sin 100tmA$ , $V = 100\sin \left( {100t + \cfrac{\pi }{3}} \right)V$ The power dissipated in the circuit is

  1. 10 W

  2. 2.5 W

  3. 5 W

  4. 5 kW
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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