Physics

Electrical Machines and Circuits

720 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 electro-magnetism applications of electromagnets transformer transformers

A transformer is used to

  1. convert DC into AC

  2. convert AC into DC

  3. obtain the required DC voltage

  4. obtain the required AC voltage
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A transformer is used to change the voltage low to high or vice versa depending upon the efficiency for power transfer and usage and since it is an application of the changing emf due to change in magnetic flux which is created by changing current it can be only applied to a.c. .

Multiple choice physics electro-magnetism applications of electromagnets transformer transformers

Which of the remain constant in a transformer?

  1. current

  2. potential

  3. power

  4. frequency
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A transformer changes Voltages and current but the power remains constant but that is case only for ideal transformer, in general there is loss in power in form of heat , sound , vibrations, eddy currents , etc .Only the frequency remains constant.

Multiple choice physics electro-magnetism applications of electromagnets transformer transformers

How are the e.m.f. in the primary and secondary coils of a transformer related with the number of turns in these coils?

  1. $E _s/ E _p = (N _s / N _p)^2$
  2. $E _s E _p = N _s N _p$
  3. $E _p/ E _s = N _s / N _p$
  4. $E _s/ E _p = N _s / N _p$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In a transformer, source of alternating current is applied to the primary coil. Due to this, the current in the primary coil (called as magnetizing current) produces alternating flux in the core of transformer. This alternating flux gets linked with the secondary coil, and because of the phenomenon of mutual induction an emf gets induced in the secondary coil. 
Mutual induction:
An alternating current flowing in a coil produces alternating magnetic field around it. When two or more coils are magnetically linked to each other, then an alternating current flowing through one coil causes an induced emf across the other linked coils. This phenomenon is called as mutual induction.
From these principles, it can be derived that emf divided by number of turns is same for both primary and secondary winding.

That is,$ \dfrac { E1 }{ N1 } =\dfrac { E2 }{ N2 } $

where,
E1 is the induced EMF in the primary coil
E2 is the induced EMF in the secondary coil
N1 is the number of turns in the primary coil
N2 is the number of turns in the secondary coil

Multiple choice physics electro-magnetism applications of electromagnets transformer transformers

For what purpose are the transformers used? Can they be used with a direct current source?

  1. To step up the a.c. voltage. No, a transformer cannot be used with a d.c. source.

  2. To step down the a.c. voltage. No, a transformer cannot be used with a d.c. source.

  3. both A & B

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

Transformers are found everywhere alternating current is used. This includes both large power stations to the power cord for our laptop computer. A transformer is an electrical device that trades voltage for current in a circuit, while not affecting the total electrical power. This means it takes high-voltage electricity with a small current and changes it into low-voltage electricity with a large current, or vice-versa. That is, step up or step down the voltage in a circuit.
A step-up transformer turns low-voltage electricity into high-voltage electricity while dropping the current. A step-down transformer changes high-voltage electricity into low-voltage electricity.
For a transformer to work, the current in one coil has to somehow make current flow in the other coil and the circuit it's connected to. A DC current in one coil will make a magnetic field on the other coil, but a magnetic field by itself won't drive any electrons around. A changing magnetic field, however, does create an electric force which will accelerate those electrons in the other coil into carrying a current. This process is described by Faraday's law of induction. We get a changing field from an AC current, since the current which makes the field is changing. Hence, DC current cannot be used as a source for a generator.

Multiple choice physics electro-magnetism applications of electromagnets transformer transformers

For what purpose are the transformers used ?

  1. To step down the a.c. voltage.

  2. To step up the a.c. voltage.

  3. both A & B

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

This question is already present in ID: 177624.

Transformers are found everywhere alternating current is used. This includes both large power stations to the power cord for our laptop computer. A transformer is an electrical device that trades voltage for current in a circuit, while not affecting the total electrical power. This means it takes high-voltage electricity with a small current and changes it into low-voltage electricity with a large current, or vice-versa. That is, step up or step down the voltage in a circuit. 
A step-up transformer turns low-voltage electricity into high-voltage electricity while dropping the current. A step-down transformer changes high-voltage electricity into low-voltage electricity. 
For a transformer to work, the current in one coil has to somehow make current flow in the other coil and the circuit it's connected to. A DC current in one coil will make a magnetic field on the other coil, but a magnetic field by itself won't drive any electrons around. A changing magnetic field, however, does create an electric force which will accelerate those electrons in the other coil into carrying a current. This process is described by Faraday's law of induction. We get a changing field from an AC current, since the current which makes the field is changing. Hence, DC current cannot be used as a source for a generator.

Multiple choice physics electro-magnetism applications of electromagnets transformer transformers

The input and output voltages of a transformer are $220$ V and $44$ V  respectively. Find the turns ratio.

  1. $2 : 5$
  2. $1 : 2$
  3. $1 : 5$
  4. $5 : 1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A transformer takes high-voltage electricity with a small current and changes it into low-voltage electricity with a large current, or vice versa. That is, step up or step down the voltage in a circuit. A step-up transformer turns low-voltage electricity into high-voltage electricity while dropping the current. A step-down transformer changes high-voltage electricity into low-voltage electricity.
If we change the number of turns in the coils we change the induced emf. This allows you to change (transform) the voltage from the primary to the secondary coil.

The Turns Rule is: $\dfrac { Ns }{ Np } =\dfrac { Vs }{ Vp } $

Where,
$Ns =$ number of turns on the secondary coil
$Np =$ number of turns on the primary coil
$Vs =$ voltage across the secondary coil
$Vp =$ voltage across the primary coil
Hence, the ratio of input voltage to output voltage is equal to the ratio of the number of turns of primary coil to the number of turns of the secondary coil.
In the question given, the ratio of the input-output voltage is $\dfrac { 220V }{ 44V } $. That is, $1:5$.

Therefore, the turns ratio will also be $1:5$.

Multiple choice physics electro-magnetism applications of electromagnets transformer transformers

A transformer is used to light a $140W,24V$ bulb from a $240V$ a.c mains. The current in the main cable is $0.7A$. The efficiency of the transformer is

  1. $63.8$%
  2. $83.3$%
  3. $16.7$%
  4. $36.2$%
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\text{Power source}=EI=240\times 0.7=166$
$\Rightarrow \eta =\dfrac{\text{Output}}{\text{Input}}=\cfrac{140}{166}$ $\Rightarrow \eta=83.3\%$

Multiple choice physics electro-magnetism applications of electromagnets transformer transformers

Statement A: Step up transformer converts low voltage, high current to high voltage, low current.
Statement B: Transformer works on both ac and dc.

  1. A is true but B is false

  2. Both A and B are true

  3. A is false but B is true

  4. Both A and B are false
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Step up transformer increases the voltage in secondary coil and decreases current in secondary coil. Transformer works only on ac current because in DC current, there is no change of flux associated with the coil

Multiple choice physics electro-magnetism applications of electromagnets transformer transformers

Maximum efficiency of a transformer depends on :

  1. the working conditions of technicians
  2. whether copper loss $=$1/2 x iron loss
  3. whether copper loss $=$ iron loss
  4. whether copper loss $=$2 x iron loss
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The efficiency of the transformer is defined as ratio of useful power output to the input power. The two being measured is the same unit. Its unit is either watts or kw. Transformer efficiency is denoted by $\eta$

$\eta = \dfrac{\text{output power}}{\text{input power}} = =\dfrac{\text{output power}}{\text{input power + losses}}$

$\eta  = \dfrac{\text{output power}}{\text{input power + iron losses of copper + losses}}$

$\eta = \dfrac{V _2I _2 \cos \phi _2}{V _2 I _2 \cos \phi _2 + P _i + P _c}$

Where

$V _2  = $ secondary terminal voltage

$I _2 = $ Full load secondary current.

$\cos \phi _2 = $ Power factor of the load

$P _i = $ Iron losses = constant loss = exactly current loss + hysteresis loss

$P _c = $ full load copper loss

The efficiency is a function of load i.e. lead current $I _2$ assuming $\cos \phi$ constant. The secondary terminal voltage $V _2$ is also assumed constant.

$\dfrac{d \eta}{d I _2} = 0$

Now $\eta = \dfrac{V _2I _2\cos \phi _2}{ V _2I _2\cos \phi _2 + P _i + I _2^2 R _{2e} }$

$\dfrac{d \eta}{d I _2} = \dfrac{d}{d I _2}(\dfrac{V _2I _2\cos \phi _2}{ V _2I _2\cos \phi _2 + P _i + I _2^2 R _{2e} }) = 0$

$( V _2I _2\cos \phi _2 + P _i + I _2^2 R _{2e}) \dfrac{d}{d I _2}( V _2I _2\cos \phi _2) – (V _2I _2\cos \phi _2) \dfrac{d}{d I _2}( V _2I _2\cos \phi _2 + P _i + I _2^2 R _{2e}) = 0$

$( V _2I _2\cos \phi _2 + P _i + I _2^2 R _{2e})(V _2 \cos \phi _2) – (V _2 I _2 \cos \phi _2)(V _2 \cos \phi _2 + 2 I _2 R _{2e}) = 0$

Cancelling $V _2 \cos \phi _ 2$ from both the terms we get

$ V _2I _2\cos \phi _2 + P _i + I _2^2 R _{2e} – V _2I _2\cos \phi _2 – 2I^2 R _{2e} = 0 $

$P _i – I^2 R _{2e} = 0$

$P _i = I^2 R _{2e} = P _{Cu}$

$P _i = P _{Cu}$

 So, condition to achieve maximum efficiency is that

Copper Loss = Iron Loss

Multiple choice physics electro-magnetism applications of electromagnets transformer transformers

A 220 - volt input is supplied to a Transformer. The output circuit draws a current of 2.0 ampere at 440 volts. If the efficiency of the transformer is 80%, the current drawn by the primary windings of the transformer is

  1. 5.0 ampere

  2. 3.6 ampere

  3. 2.8 ampere

  4. 2.5 ampere
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\mu =\frac { { V } _{ s }{ I } _{ s } }{ { V } _{ p }{ I } _{ p } } =0.8\Rightarrow { I } _{ p }=\frac { { V } _{ s }{ I } _{ s } }{ { V } _{ p }\mu  } =\frac { 440\times 2 }{ 220\times .8 } =5A$

Multiple choice physics electro-magnetism applications of electromagnets transformer transformers

Electrical energy is transmitted over large distance at high alternating voltages Which of the following statements is (are) correct?

  1. for a given power level there is a lower current

  2. lower current implies less power loss

  3. transmission lines can be made thinner

  4. It is easy to reduce the voltage at the receiving end using step-down transformers
Reveal answer Fill a bubble to check yourself
A,B,D Correct answer
Explanation

Power = Voltage x Current
As voltages are high so currents are lower. This is done to reduce the resistive losses.
Resistance Loss = ${ { I } _{ rms }^{ 2 } }R$
So, lower current implies less power loss. 
As resistance is inversely proportional to their diameter; hence decreasing their diameter by x will increase resistance by $x^2$ and hence increasing power loss hence not done.
As the voltage required for domestic and commercial uses are pretty less then they can be easily brought down by step-down transformers. 

Multiple choice physics electro-magnetism applications of electromagnets transformer transformers

Tum ratio in a step up transformer is 1 : 2 if a Lechlanche cell of 1.5 V is connected across the input, then the voltage across the output will be 

  1. 0.1 V

  2. 1.5 V

  3. 0.75 V

  4. Zero
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A transformer can not step up a d.c. input so output potential here will be zero. No potential will be induced in the secondary coil.

Multiple choice physics electro-magnetism applications of electromagnets transformer transformers

Large transformers, when used for some time, become very hot and are cooled by circulating oil. The heating of the transformer is due to

  1. The heating effect of current alone

  2. Hysteresis loss alone

  3. Both the heating effect of current and hysteresis loss

  4. Intense sunlight at noon
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation 

Heat loss in a magnetic circuit is due to two reasons one is hysteresis and  other is due to ohmic loss due to eddy current .Laminations and circulatingoil are some of methods to avoid heating. Hysteresis loss is entirelymaterial property while eddy current loss depends on geometry of core and amount of current.

Multiple choice physics electro-magnetism applications of electromagnets transformer transformers

A certain transformer has a turns ratio of 1 and a 0.85 coefficient of coupling. When 2 V ac is applied to the primary, the secondary voltage is

  1. 1.7 V

  2. 0.85 V

  3. 1 V

  4. 0 V
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Coefficient coupling  $\eta = 0.85$
Input voltage in primary coil   $V _p = 2 \ V$
Turn ratio  $\dfrac{N _s}{N _p} = 1$
Thus output voltage in secondary coil   $V _s = \eta \dfrac{N _s}{N _p}V _p$
$\implies \ V _s = 0.85\times 1\times 2 = 1.7 \ V$