Tag: alternating current

Quizzes (2) Questions (29)

Questions Related to alternating current

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

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. $1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

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. $1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice physics alternating current power in ac circuits average power in ac circuit and power factor power in ac circuit

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. $\dfrac{E^{2} _{0}}{8R}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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.

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

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

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).

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.