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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

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. $70 V$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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.

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

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. $55.5\ rad/s, 90$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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.

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

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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 option c: imaging optical instruments : telescope the reflecting telescope optical telescope xx radio telescope and space telescopes

A reflecting telescope has a large mirror for its objective with radius of curvature equal to 80 cm. The magnifying power of this telescope if eye piece used has a focal length of 1.6 cm is:

  1. 100

  2. 50

  3. 25

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

The focal length of objective mirror $f _0 = \dfrac{R}{2} = \dfrac{80}{2} = 40 \,cm$
and focal length of eye piece = 1.6 cm
$\therefore $ magnifying power, $m = \dfrac{f _0}{f _e} = \dfrac{40}{1.6 } = 25$