Physics

Electrical Machines and Circuits

754 Questions

Electrical machines and circuits involve the principles of alternating and direct current, transformers, diodes, and power consumption. These concepts are essential for engineering and technical exams. Review these practice questions to test your knowledge of circuit analysis.

Transformers and impedanceAlternating and direct currentCircuit components and diodesPower consumption calculations

Electrical Machines and Circuits Questions

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

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

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 some natural phenomena lightning conductors lightning safety advanced of lightning


Two types of earthings are provided: (a) Equipment earthing and (b) Neutral point earthing. Equipment earthing is provided for

  1. Safety of human being

  2. Safety of equipments

  3. Safety of starters and cables

  4. None
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Equipment earthing is a connection done through a metal link between the body of any electrical appliance, or neutral point, as the case may be, to the deeper ground soil. The metal link is normally of MS flat, CI flat, GI wire which should be penetrated to the ground earth grid.

  • Safety of personnel
  • Safety of equipment Prevent or at least minimize damage to equipment as a result of flow of heavy currents.
  • Improvement of the reliability of the power system

Option A is correct.

Multiple choice physics some natural phenomena lightning conductors lightning safety advanced of lightning

Which of the following correctly represents the sequence of operations of isolator circuit breaker and earthing switch while opening a circuit :

  1. close earthing switch - open circuit breaker open isolator

  2. open isolator - close circuit breaker - open earthing switch

  3. open circuit breaker - open isolator - close earthing switch

  4. close circuit breaker - close isolator - open earthing switch
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Open circuit breaker - open isolator - close earthing switch represents the sequence of operations of isolator circuit breaker and earthing switch while opening a circuit.


Option C is correct.

Multiple choice physics some natural phenomena lightning conductors lightning safety advanced of lightning


Which of the following correctly presents the sequence of operations of isolator circuit breaker and earthing switch while closing a circuit :

  1. Ensure circuit breaker is closed - close isolator - open earthing switch

  2. Ensure circuit breaker is open - close isolator - open earthing switch if any close circuit breaker

  3. Ensure circuit breaker is open - open isolator - open earthing switch if any close circuit breaker

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

 Ensure circuit breaker is open - close isolator - open earthing switch if any close circuit breaker .


Option B is correct.

Multiple choice physics current electricity meter bridge meter bridge and problems on it galvanometer

A lamp of 6 V and 30 W is used in a laboratory but the supply is of 120 V. what will be done to make use of the lamp?
(1) A resistance may be used
(2) A resistance may be used in series with lamp.
(3) The resistance should be of 18 $\Omega$. 

  1. 1, 2 and 3 are correct

  2. 1 and 2 are correct

  3. 1 and 3 are correct

  4. 2 and 3 are correct
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Here, the resistance should be used in series with lamp in order to divide the voltage or to reduce potential drop across the lamp.

Lamp can withstand a maximum of 6 V , and a current of $\dfrac { 6 }{ { R } _{ lamp } } $ where ${R} _{lamp}$ is $\dfrac { { 6 }^{ 2 } }{ 30 } =1.2\Omega $
hence maximum current that lamp can with stand is  $\dfrac { 6 }{ { 1.2} }= 5A $
so the resistance which is to be connected in series would also get same current of 5 A hence if it will be of $18 \Omega$ then voltage drop accros it will be 90 V and maximum of 6 V across lamp but both voltage drop does not make 120 V . hence the resistance value should be more than $18 \Omega$.  So 1 and 2 is correct but 3 is incorrect.

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

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

Multiple choice evs sources of water conservation of natural and energy resources energy conservation and sources importance of the conservation of natural resources

Switch on the power when TV is not in use leads to the energy loss of

  1. 5 watts

  2. 10 watts

  3. 1 watts

  4. 50 watts
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The energy management is very important for the conservation of energy. It involves monitoring the consumption of energy, finding ways to save (control) energy and taking initiatives to conserve energy by replacing the inefficient equipments and controlling routine waste. The power should not be switched on when TV and any other audio or video device is not in use for a long time. It is because even in the switch off condition energy is consumed by the electric appliances and this is called standby power. 10 watts of energy is consumed by the TV even if, the power is on but TV is not in use. So, the correct option is B.