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 electrical circuits domestic electric circuits and safety precautions domestic and commercial circuit electric power

Household electrical appliances are joined using ________ combination of resistors

  1. Parallel

  2. Alternating

  3. Continuous

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

In a house, there are many electrical appliances that have to run independent of each other. If one appliance is turned on or off it should not affect the other appliances. Thus even if a single appliance is taken out of the circuit (turned off / open switched ) the circuit breaks and hence the current cannot flow in the circuit so every thing turns off.

 So even if a single appliance is taken out of the circuit (turned off / open switched ) the circuit breaks and hence the current cannot flow in the circuit .
In parallel, even if an appliance gets damaged or is turned off ,the current always has other independent dedicated parallel paths to other appliances and hence every appliance is still works well. 

Multiple choice physics electrical circuits domestic electric circuits and safety precautions domestic and commercial circuit electric power

Which statement is wrong from the following?

  1. The earthing wire is green coloured.
  2. In India, current flowing through wire is AC and its frequency is $50 \,Hz$.
  3. In India, voltage between two wires is $110 \,V$.
  4. T.V., Tublight, bulbs are connected with $5\, A$ line.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In India, current flowing through wire is A.C and its frequency is $50Hz$ and voltage $220V$.But, Option C said voltage between two wires is 110V110V, which is wrong.

Therefore, C is correct option.

Multiple choice physics electrical circuits domestic electric circuits and safety precautions domestic and commercial circuit electric power

Which of the following statements is true?

  1. In tree type distribution of electric power, fuses are present only on the main board.

  2. In ring type distribution of electric power, there is an individual fuse for each appliance.

  3. Ring type distribution of electric power is advantageous than the tree type distribution.

  4. All the above.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In tree type distribution of electric power, fuses are present only on the distribution board where as in ring type, there is an individual fuse for each appliance. As there is an individual fuse for each appliance in ring type, only faulty appliance will not work and all the other appliances will not be affected. It is easier to install. So, ring type is advantageous than that of tree type. Hence, choice is (4)

Multiple choice physics electrical circuits domestic electric circuits and safety precautions domestic and commercial circuit electric power

What is the frequency of 220$\mathrm { v } \mathrm { DC }$ voltage? 

  1. Zero HZ
  2. 50$\mathrm { Hz }$
  3. 60$\mathrm { Hz }$
  4. 220$\mathrm { Hz }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Direct Current (DC) is characterized by a constant flow of charge in one direction, meaning it does not oscillate. Therefore, its frequency is zero.

Multiple choice physics electrical circuits domestic electric circuits and safety precautions domestic and commercial circuit electric power

A geyser is rated 1500 W, 250 V. This geyser is connected to 250 V mains. The current drawn will be:

  1. $6 A$
  2. $5 A$
  3. $40 A$
  4. $10 A$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Power, $P=VI$
For a $1500\ W$ geyser rated for $250\ V$, the current through is given as $I=\dfrac { P }{ V } =\dfrac { 1500 }{ 250 } =6A$
Hence, the current flowing through the heater is $6\ A$.

Multiple choice physics electrical circuits domestic electric circuits and safety precautions domestic and commercial circuit electric power

Three heaters each rated 250 W, 100 V are connected in parallel to a 100 V supply. The total current taken from the supply is :

  1. $2.5 A$
  2. $5 A$
  3. $7.5 A$
  4. $25 A$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Power is given as: $P=VI$


For a $60\ W$ lamp rated for $ 250\ V$, the current through is given as: $I=\dfrac { P }{ V } =\dfrac { 250 }{ 100 } =2.5\ A$

Hence, the current flowing through the heater is $2.5\ A$.

In a parallel circuit, the voltage across each of the components is the same, and the total current is the sum of the currents through each component.
Therefore, when three such heaters are connected in parallel, then the current obtained from the supply is given as $2.5+2.5+2.5 = 7.5\ A$

Multiple choice physics electrical circuits domestic electric circuits and safety precautions domestic and commercial circuit electric power

The current through a 60 W lamp rated for 250 V is 

  1. 0.24 A

  2. 4.2 A

  3. 0.5 A

  4. 6 A
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The electric power in watts associated with a complete electric circuit or a circuit component represents the rate at which energy is converted from the electrical energy of the moving charges to some other form, e.g., heat, mechanical energy, or energy stored in electric fields or magnetic fields. The power is given by the product of applied voltage and the electric current. That is, $P=VI$. 

For a 60 W lamp rated for 250 V, the current through is given as $I=\dfrac { P }{ V } =\dfrac { 60 }{ 250 } =0.24A$.

Hence, the current flowing through the lamp is 0.24 A.

Substituting I=V/R in the above formula, we get, $P=\dfrac { { V }^{ 2 } }{ R } $.

Given that the voltage is 250 V and the power is 60 W, the resistance of the 

bulb is calculated as follows.

$R=\dfrac { { V }^{ 2 } }{ P } =\dfrac { { 250 }^{ 2 } }{ 60 } =1041.67\Omega$.

When the voltage drops to 200 V, the power is calculated as follows.

$P=\dfrac { { V }^{ 2 } }{ R } =\dfrac { { 200 }^{ 2 } }{ 1041.67 } =\quad 38.40W$

Hence, the power of the bulb is reduced to 38.40 W.
Multiple choice physics electrical circuits domestic electric circuits and safety precautions domestic and commercial circuit electric power

An electric heater consists of a nichrome coil and runs under $220 V$, consuming $1 kW$ power. Part of its coil burned out and it was reconnected after cutting off the burnt portion, The power it will consume now is:

  1. More than $1 kW$
  2. Less that $1 kW$, but not zero
  3. $1 kW$
  4. $0kW$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Answer is A.

The electrical resistance of a wire would be expected to be greater for a longer wire, less for a wire of larger cross sectional area, and would be expected to depend upon the material out of which the wire is made.The resistance of a wire can be expressed as $R=\rho \frac { L }{ A } $, 
where,
$\rho $ - Resistivity  - the factor in the resistance which takes into account the nature of the material is the resistivity
L - Length of the conductor
A - Area of cross section of the conductor.
From this relation, we observe that the length is directly proportional to the resistance and the area of cross section is inversely proportional to the resistance.
In this case, the length of the nichrome coil is reduced due to the burn and so the resistance will also be reduced proportionally. 
When the resistance is decreased, more current flows through the coil and apparently, more power is consumed by the heater as power consumed P = VI.
Hence, the power it will consume now is more than 1 kW.

Multiple choice physics electrical circuits domestic electric circuits and safety precautions domestic and commercial circuit electric power

A neon lamp is connected to a voltage a.c. source. The voltage is gradually increased from zero volt. It is observed that the neon flashes at $50 V$. The a.c, source is now replaced by a variable dc source and the experiment is repeated. The neon bulb will flash at ?

  1. $50V$
  2. $70V$
  3. $100V$
  4. $35V$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Answer is B.

The RMS (Root mean square) value tells us what equivalent DC voltage we would need to get the same power, for the neon bulb to glow.
In this case, the neon bulb glows at 50 V ac voltage. Therefore, for the bulb glow with dc voltage, $50\times \sqrt { 2 }  =50\times 1.414=70V$ dc voltage should be applied.
Hence, the neon bulb will flash at 70 V dc.