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

in electric bulb rated $160W,80\ V$ has to be operated across  $100\ V,50Hz$ a.c supply. The reactance Which is series with bulb should be-

  1. $30\ \Omega$
  2. $60\ \Omega$
  3. $15\ \Omega$
  4. $120\ \Omega$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

First find the resistance of the bulb using R = V^2 / P = 80^2 / 160 = 40 ohms, and its rated current I = P / V = 160 / 80 = 2 A. When connected across a 100 V supply, the total impedance Z of the bulb and inductor in series is Z = V_total / I = 100 / 2 = 50 ohms. Since Z^2 = R^2 + X_L^2, we get 50^2 = 40^2 + X_L^2, which gives X_L = 30 ohms.

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.

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

Two dissimilar bulbs are connected in series, which bulb will be brighter ?

  1. low resistance bulb

  2. more resistance bulb

  3. high current pass bulb.

  4. low current pass bulb
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Components connected in series are connected along a single path, so the same current flows through all of the components. The current through each of the components is the same, and the voltage across the circuit is the sum of the voltages across each component.
When two bulbs are connected in series the first bulb will receive more current. As the bulb has its own resistance, less current will flow to the next bulb.
Hence the first bulb will glow more.

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

An electric oven of 2 kW power rating is operated in a domestic electric circuit that has a current rating of 5A. If the supply voltage is 220V, what result do you expect?

  1. circuit will be breaked

  2. fuse will blow

  3. both

  4. none
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given,
Power of the oven $(P) = 2kw = 2 \times 103 W$
Voltage supplied $(V) = 220V$
Current $(I) = ? A$
Power $= V \times I$
$\Rightarrow I = P/V = 2 \times 103/220 = 9.09 A$
Since the domestic electric circuit has a current rating of 5A, the flow of 9.09 A by the Oven exceeds the safe limit. Here, fuse will blow and break the circuit.

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

What is the function of distribution box in domestic electrical circuit

  1. It distributes voltage across appliances as per need

  2. It provides isolation between set of appliances and improves reliability

  3. It distributes Losses among all parts of circuit equally

  4. It allows user to manually distribute power as per his need
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Distribution box in domestic electrical circuit provides isolation between appliances and allow them to work paralelly. It also ensures failure in one part do not affect the functioning of the other.

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

A transformer is used to light $100\ W - 110\ V$ lamp from $220\ V$ mains. If the main current is $0.5\ A$, the efficiency of the transformer is

  1. $90$%
  2. $95$%
  3. $96$%
  4. $99$%
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Voltage in primary coil     $V _p = 220$ V

Current in primary coil    $I _p = 0.5$ A
Thus input power     $P _{in} = V _pI _p$
Output power  (in secondary coil )      $P _{out} = 100$ W
$\therefore$  Efficiency of transformer     $\eta = \dfrac{P _{out}}{P _{in}}\times 100$
$\implies   $        $\eta = \dfrac{100}{(220)(0.5)}\times 100 = 90$ %.