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Capacitance and Capacitors MCQs ( Electrical Engineering ) MCQs – Competitive Electrical Engineering MCQs

Capacitance and Capacitors MCQs ( Electrical Engineering ) MCQs – Competitive Electrical Engineering MCQs 

Latest Electrical Engineering MCQs

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Capacitance and Capacitors MCQs ( Electrical Engineering ) MCQs – Competitive Electrical Engineering MCQs

The most occurred mcqs of Capacitance and Capacitors MCQs ( Electrical Engineering ) in past papers. Past papers of Capacitance and Capacitors MCQs ( Electrical Engineering ) Mcqs. Past papers of Capacitance and Capacitors MCQs ( Electrical Engineering ) Mcqs . Mcqs are the necessary part of any competitive / job related exams. The Mcqs having specific numbers in any written test. It is therefore everyone have to learn / remember the related Capacitance and Capacitors MCQs ( Electrical Engineering ) Mcqs. The Important series of Capacitance and Capacitors MCQs ( Electrical Engineering ) Mcqs are given below:

Charging and Discharging Currents

1. Which of the following depends on charging and discharging rate of a capacitor?
a) Time constant
b) Current
c) Power
d) Voltage
Answer: a
Explanation: The time constant in a circuit consisting of a capacitor is the product of the resistance and the capacitance. Smaller the time constant, faster is the charging and discharging rate and vice versa.


2. What is the initial current while charging a capacitor?
a) High
b) Low
c) 0
d) Cannot be determined
Answer: a
Explanation: The initial current of a capacitor is very high because the voltage source will transport charges from one plate of the capacitor to the other plate.


3. What is the final current while charging a capacitor?
a) High
b) Zero
c) Infinity
d) Low
Answer: b
Explanation: The final current is almost equal to zero while charging a capacitor because the capacitor is charged up to the source voltage.


4. A capacitor is charged to a voltage of 400V and has a resistance of 20ohm. Calculate the initial value of charging current.
a) 10A
b) 0A
c) Infinity
d) 20A
Answer: d
Explanation: When the capacitor is charging the initial value if the current is V/R=400/20 = 20A.


5. A capacitor is charged to a voltage of 400V and has a resistance of 20ohm. Calculate the initial value of the discharge current.
a) 10A
b) 0A
c) Infinity
d) 20A
Answer: b
Explanation: When the capacitor is discharging the value of the initial current is zero.


6. A capacitor is charged to a voltage of 400V and has a resistance of 20ohm. Calculate the final value of the discharge current.
a) 10A
b) 0A
c) Infinity
d) 20A
Answer: d
Explanation: In a discharging circuit, the final voltage is equal to zero for capacitor. For a resistor, final voltage is 400V.So,final current = V/R = 400/20 = 20A.


7. When will be capacitors fully charged?
a) When voltage is zero
b) When the supply voltage is equal to the capacitor voltage
c) When voltage is infinity
d) When capacitor voltage is equal to half the supply voltage
Answer: b
Explanation: When the capacitor voltage is equal to the source voltage, it means that all the charges have moved from one plate of the capacitor to the other.


8. What happens to the capacitor when the capacitor voltage is equal to the source voltage?
a) The charging phase of the capacitor is over
b) The discharging phase of the capacitor is over
c) The capacitor is switched off
d) The capacitor is switched on
Answer: c
Explanation: When the capacitor voltage is equal to the source voltage, it means that all the charges have moved from one plate of the capacitor to the other. Hence the capacitor is fully charged and we say it gets switched off.


9. A capacitor is charged to a voltage of 400V and has a resistance of 20ohm. Calculate the final value of charging current.
a) 10A
b) 0A
c) Infinity
d) 20A
Answer: b
Explanation: When the capacitor is charging, the final voltage of the capacitor becomes equal to the voltage of source. Hence, the current becomes equal to zero.

Growth and Decay

1. The charging time constant of a circuit consisting of a capacitor is the time taken for the charge in the capacitor to become __________ % of the initial charge.
a) 33
b) 63
c) 37
d) 36
Answer: b
Explanation: We know that: Q=Q0(1-e-t/RC).
When RC=t, we have: Q=Q0(1-e-1) = 0.63*Q0.
Hence the time constant is the time taken for the charge in a capacitive circuit to become 0.63 times its initial charge.


2. The discharging time constant of a circuit consisting of a capacitor is the time taken for the charge in the capacitor to become __________ % of the initial charge.
a) 33
b) 63
c) 37
d) 36
Answer: c
Explanation: We know that: Q=Q0(1-e-t/RC).
When RC=t, we have: Q=Q0(1-e-1) = 0.37*Q0.
Hence the time constant is the time taken for the charge in a capacitive circuit to become 0.37 times its initial charge.


3. A circuit has a resistance of 2 ohm connected in series with a capacitance of 6F. Calculate the charging time constant.
a) 3
b) 1
c) 12
d) 8
Answer: c
Explanation: The charging time constant in a circuit consisting of a capacitor and resistor in series is the product of the resistance and capacitance = 2*6 = 12.


4. A circuit has a resistance of 5 ohm connected in series with a capacitance of 10F. Calculate the discharging time constant.
a) 15
b) 50
c) 5
d) 10
Answer: b
Explanation: The discharging time constant in a circuit consisting of a capacitor and resistor in series is the product of the resistance and capacitance = 5*10 = 50.


5. What is the value of current in a discharging capacitive circuit if the initial current is 2A at time t=RC.
a) 0.74A
b) 1.26A
c) 3.67A
d) 2.89A
Answer: b
Explanation: At time t=RC, that is the time constant, we know that the value of current at that time interval is equal to 63% of the initial charge in the discharging circuit. Hence, I = 2*0.63 = 1.26A.


6. What is the value of current in a charging capacitive circuit if the initial current is 2A at time t=RC.
a) 0.74A
b) 1.26A
c) 3.67A
d) 2.89A
Answer: a
Explanation: At time t=RC, that is the time constant, we know that the value of current at that time interval is equal to 37% of the initial charge in the discharging circuit. Hence, I = 2*0.37 = 0.74A.


7. While discharging, what happens to the current in the capacitive circuit?
a) Decreases linearly
b) Increases linearly
c) Decreases exponentially
d) Increases exponentially
Answer: d
Explanation: The equation for the value of current in a discharging capacitive circuit is:
I=I0(1-e-t/RC). From this equation, we can see that the current is exponentially increasing.


8. While discharging, what happens to the voltage in the capacitive circuit?
a) Decreases linearly
b) Increases linearly
c) Decreases exponentially
d) Increases exponentially
Answer: c
Explanation: The equation for the value of voltage in a discharging capacitive circuit is:
V=V0(e-t/RC). From this equation, we can see that the voltage is exponentially decreasing.


9. While charging, what happens to the current in the capacitive circuit?
a) Decreases linearly
b) Increases linearly
c) Decreases exponentially
d) Increases exponentially
Answer: c
Explanation: The equation for the value of current in a charging capacitive circuit is:
I=I0(e-t/RC). From this equation, we can see that the current is exponentially decreasing.


10. While charging, what happens to the voltage in the capacitive circuit?
a) Decreases linearly
b) Increases linearly
c) Decreases exponentially
d) Increases exponentially
Answer: d
Explanation: The equation for the value of voltage in a charging capacitive circuit is:
V=V0(1-e-t/RC). From this equation, we can see that the voltage is exponentially increasing.

Discharge of a Capacitor Through a Resistor

1. An 8microF capacitor is connected in series with a 0.5 megaohm resistor. The DC voltage supply is 200V. Calculate the time constant.
a) 1s
b) 2s
c) 3s
d) 4s
Answer: d
Explanation: The time constant is the product of the resistance and capacitance in a series RC circuit.
Therefore, time constant = 8*10-6*4*106=4s.


2. An 8microF capacitor is connected in series with a 0.5 megaohm resistor. The DC voltage supply is 200V. Calculate the initial charging current.
a) 100 microA
b) 500 microA
c) 400 microA
d) 1000microA
Answer: c
Explanation: In a series RC circuit, the initial charging current is:
I=V/R = 200/(0.5*106s) = 400*10-6A = 400 microA.


3. A 8 microF capacitor is connected in series with a 0.5 megaohm resistor. The DC voltage supply is 200V. Calculate the time taken for the potential difference across the capacitor to grow to 160V.
a) 6.93s
b) 7.77s
c) 2.33s
d) 3.22s
Answer: a
Explanation: From the previous explanations, we know that the initial current is 400mA and the time constant is 4s. Substituting the values of capacitor voltage, initial voltage, initial current and time constant in the equation: V=V0(1-e-t/RC)
Substituting V=160V, V0=200V, RC=4s we get,
t=6.93s.


4. An 8microF capacitor is connected in series with a 0.5 megaohm resistor. The DC voltage supply is 200V. Calculate the voltage in the capacitor 4s after the power is supplied.
a) 123.4V
b) 126.4V
c) 124.5V
d) 132.5V
Answer: b
Explanation: We can get the value of the potential difference across the capacitor in 4s, from the following equation:
Vc=V(1-e-t /RC). Substituting the values in the given equation, we get Vc = 126.4V.


5. An 8microF capacitor is connected in series with a 0.5 megaohm resistor. The DC voltage supply is 200V. Calculate the current in the capacitor 4s after the power is supplied.
a) 79 microA
b) 68 microA
c) 48 microA
d) 74 microA
Answer: d
Explanation: In the given question, the time constant is equal to the time taken= 4s. Hence the value of current will be 37% of its initial value = I=0.37*200 = 74 microA.

 

Simple Magnetic Circuits MCQs




6. A circuit has a resistance of 2 ohms connected in series with a capacitance of 6F. Calculate the discharging time constant.
a) 3
b) 1
c) 12
d) 8
Answer: c
Explanation: The discharging time constant in a circuit consisting of a capacitor and resistor in series is the product of the resistance and capacitance = 2*6 = 12.


7. What is the energy in a capacitor if the voltage is 5V and the charge is10C?
a) 25J
b) 35J
c) 54J
d) 55J
Answer: a
Explanation: We know that Q/V=C. Hence the value of capacitance is 2F.
U=(Q*V)/2 = (10*5)/2 = 25 J.

Transients in CR Networks

1. A CR network is one which consists of _________
a) A capacitor and resistor connected in parallel
b) A capacitor and resistor connected in series
c) A network consisting of a capacitor only
d) A network consisting of a resistor only
Answer: b
Explanation: A CR network is one which consists of a capacitor connected in series with a resistor. The capacitor discharges or charges through the resistor.


2. At DC, capacitor acts as _________
a) Open circuit
b) Short circuit
c) Resistor
d) Inductor
Answer: a
Explanation: Capacitive Reactance XC = 1/(2πfC)
For DC, f=0 so, XC becomes infinite. Hence for dc, the capacitor acts as an open circuit.


3. In an RC series circuit, when the switch is closed and the circuit is complete, what is the response?
a) Response does not vary with time
b) Decays with time
c) Increases with time
d) First increases, then decrease
Answer: b
Explanation: In an RC series circuit, the response decays with time because according to the equation, there is an exponential decrease in the response.


4. If the switch is closed at t=0, what is the current in the circuit?
basic-electrical-engineering-questions-answers-transients-cr-networks-q4
a) 0A
b) 10A
c) 20A
d) Infinity
Answer: b
Explanation: As soon as the switch is closed at t=0, the capacitor acts as a short circuit. The current in the circuit is:
I=V/R = 100/10 = 10A.


5. Calculate the voltage across the capacitor at t=0.
basic-electrical-engineering-questions-answers-transients-cr-networks-q4
a) 0V
b) 10V
c) 20V
d) Infinity
Answer: a
Explanation: When the switch is closed at t=0, the capacitor has no voltage across it since it has not been charged. The capacitor acts as a short circuit and the voltage across it is zero.


6. Calculate di(0)/dt if the switch is closed at t=0.
basic-electrical-engineering-questions-answers-transients-cr-networks-q4
a) -9.9A/s
b) -10A/s
c) 0A/s
d) -0.1A/s
Answer: d
Explanation: Applying KVL to the given circuit, we get:
i=i0e-t/RC = (100/10)e-t/100
i=10 e-t/100
di/dt = -(10/100) e-t/100
di(0)/dt=-0.1A/s.


7. Calculate d2i(0)/dt2 from the given circuit.
basic-electrical-engineering-questions-answers-transients-cr-networks-q4
a) 10-6A/s2
b) 10-3A/s2
c) 106A/s2
d) 103A/s2
Answer: b
Explanation: Applying KVL to the given circuit, we get:
100+10i(0)+1/10*integral(i(0)dt)=0
Differentiating once, we get:
10di(0)/dt+1/10*i.
Differentiating once again, we get:
10d2i(0)/dt2+10di(0)/dt=0.
Substituting the values of di/dt from the previous explanation, we get d2i(0)/dt2=10-3A/s2.


8. The current equation for the given circuit is?
basic-electrical-engineering-questions-answers-transients-cr-networks-q4
a) i=10e(-0.01)t A
b) i=10e(0.01)t A
c) i=10e(-0.001)t A
d) i=100e(-0.01)t A
Answer: a
Explanation: The KVL equation is:
100+10i(0)+1/10*integral(i(0)dt)=0
On applying Laplace transform to this equation, we get:
100/s=I(s)/10s+10I(s)
Solving the equation, we get:
i=10e(-0.01)t A.


9. The expression for the current in an RC circuit is?
a) i=(V/R)et/RC
b) i=(V/R)e-t/RC
c) i=(V/R)(1-e-t/RC)
d) i=(V/R) (1-et/RC)
Answer: b
Explanation: Applying KVL to the given circuit, we get:
i=i0e-t/RC = (100/10)e-t/100
i=10 e-t/100.


10. What is the voltage in the resistor as soon as the switch is closed at t=0.
basic-electrical-engineering-questions-answers-transients-cr-networks-q10
a) 0V
b) Infinity
c) 220V
d) Insufficient information provided
Answer: c
Explanation: As soon as the switch is closed at t=0, there is no charge in the capacitor, hence the voltage across the capacitor is zero and all the 220V voltage is the voltage across the resistor.

Energy Stored in a Charged Capacitor

1. Work done in charging a capacitor is ____________
a) QV
b) 12QV
c) 2QV
d) QV2
Answer: b
Explanation: We know that work done= Q2/2C.
Substituting C as Q/V, we get work done = Q/2V.


2. Energy stored in 2000mF capacitor charged to a potential difference of 10V is?
a) 100J
b) 200J
c) 300J
d) 400J
Answer: a
Explanation: From the expression:
WD = CV2/2 = 100J.


3. When do we get maximum energy from a set of capacitors?
a) When they are connected in parallel
b) When they are connected in series
c) Both in series and parallel
d) Insufficient information provided
Answer: a
Explanation: We get maximum energy when capacitors are connected in parallel because the equivalent capacitance is larger than the largest individual capacitance when connected in parallel. The relation between capacitance and energy is:
Energy=CV2/2, hence as the capacitance increases, the energy stored in it also increases.


4. If the charge stored in a capacitor is 4C and the value of capacitance is 2F, calculate the energy stored in it.
a) 2J
b) 4J
c) 8J
d) 16J
Answer: b
Explanation: The expression for finding the value of energy is:
U=Q2/2C = 4*4/(2*2) = 4J.


5. If the charge in a capacitor is 4C and the energy stored in it is 4J, find the value of capacitance.
a) 2F
b) 4F
c) 8F
d) 16F
Answer: a
Explanation: The expression for finding the value of energy is:
U=Q2/2C.
Substituting the values of U and Q, we get C=2F.


6. If the charge in a capacitor is 4C and the energy stored in it is 4J, calculate the voltage across its plates.
a) 2V
b) 4V
c) 8V
d) 16V
Answer: a
Explanation: The expression for finding the value of energy is:
U=Q2/2C.
Substituting the values of U and Q, we get C=2F.
V=Q/C, hence V=4/2=2V.


7. Calculate the energy in the 2F capacitor.
basic-electrical-engineering-questions-answers-energy-charged-capacitor-q7
a) 8.6kJ
b) 64kJ
c) 64J
d) 6.4kJ
Answer: d
Explanation: From the expression:
WD= CV2/2 = 2*802/2=6400J=6.4kJ.


8. Calculate the energy in the 4F capacitor.
basic-electrical-engineering-questions-answers-energy-charged-capacitor-q7
a) 128kJ
b) 1.28kJ
c) 12.8kJ
d) 1280J
Answer: c
Explanation: From the expression:
WD = CV2/2 = 4*802/2 = 12800J = 12.8kJ.


9. Calculate the energy stored in the combination of the capacitors.
basic-electrical-engineering-questions-answers-energy-charged-capacitor-q7
a) 192kJ
b) 1.92kJ
c) 19.2kJ
d) 1920J
Answer: c
Explanation: The equivalent capacitance is: Ceq=4+2=6F.
From the expression:
WD = CV2/2 = 6*802/2 = 19200J = 19.2kJ.

Force of Attraction Between Oppositely Charged Plates

1. Which among the following is the correct expression for force between the plates of a parallel plate capacitor?
a) F=epsilon*A*(V/x)2/2
b) F=epsilon*A*(V/x)2/3
c) F=epsilon (V/x)2/2
d) F=epsilon (V/x)2/3
Answer: a
Explanation: The force is proportional to the square of the potential gradient and the area. Hence the force F=epsilon*A*(V/x)2/2.


2. When the area of cross section of the plate increases, what happens to the force between the plates?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero
Answer: a
Explanation: The force of attraction between the two plates of the capacitor is directly proportional to the area of cross section of the plates, hence an area of cross section increases, the force of attraction also increases.


3. When the potential gradient increases, what happens to the force between the plates?
a) Increases
b) Decreases
c) Remains the same
d) becomes zero
Answer: a
Explanation: The force of attraction between the two plates of the capacitor is directly proportional to the square of potential gradient, hence as a potential gradient increases, the force of attraction also increases.


4. In which of the following mediums, will the force of attraction between the plates of a capacitor be greater?
a) Air
b) Water
c) Does not depend on the medium
d) Cannot be determined
Answer: b
Explanation: The absolute permittivity(epsilon) of water is greater than that of air. The expression relating F and epsilon is F=epsilon*A*(V/x)2/2. From this expression, we can see that as epsilon increases, the force of attraction also increases.


5. A metal parallel plate capacitor has 100mm diameter and the distance between the plates is 1mm. The capacitor is placed in air. Calculate the force on each plate if the potential difference between the plates is 1kV.
a) 350N
b) 0.035kN
c) 0.035N
d) 3.35kN
Answer: c
Explanation: From the given data:
A=pi*d2/4=0.007854m2
Potential gradient = V/x = 106V/m
F=epsilon*A*(V/x)2/2
Therefore, F=0.035N.


6. A metal parallel plate capacitor has 100mm diameter and the distance between the plates is ‘a’ mm. The capacitor is placed in air. Force on each plate is 0.035N and the potential difference between the plates is 1kV. Find ‘a’.
a) 1m
b) 1cm
c) 10cm
d) 1mm
Answer: d
Explanation: From the given data:
A=pi*d2/4=0.007854m2
Potential gradient = V/x = 1000/a
F=epsilon*A*(V/x)2/2
Substituting the given values, we find a=1mm.


7. A metal parallel plate capacitor has ‘a’mm diameter and the distance between the plates is 1mm. The capacitor is placed in air. Force on each plate is 0.035N and the potential difference between the plates is 1kV. Find ‘a’.
a) 10mm
b) 100mm
c) 1000m
d) 1000cm
Answer: b
Explanation: From the given data:
A=pi*d2/4=pi*a2/4
Potential gradient = V/x = 106V/m
F=epsilon*A*(V/x)2/2
Substituting the given values, we get d=100mm.


8. A metal parallel plate capacitor has 100mm diameter and the distance between the plates is 1mm. The capacitor is placed in air. Calculate the potential difference between the plates if the force on each plate is 0.035N.
a) 1kV
b) 1V
c) 2kV
d) 2V
Answer: a
Explanation: From the given data:
A=pi*d2/4=0.007854m2
Potential gradient = V/x = 1000*V
F=epsilon*A*(V/x)2/2
Substituting the given values in the above expression, we get V=1kV.

 

Electromagnetism MCQs




9. What happens to the force of attraction between the capacitors when the potential difference between the plates decreases?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero
Answer: b
Explanation: The force of attraction between the two plates of the capacitor is directly proportional to the square of the potential difference between the plates, hence as the potential difference decreases, the force of attraction also decreases.


10. What happens to the force of attraction between the capacitors when the distance of separation between the plates increases?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero
Answer: b
Explanation: The force of attraction between the two plates of the capacitor is inversely proportional to the square of the distance between the plates, hence as distance increases, the force of attraction decreases.

Dielectric Strength

1. The unit for dielectric strength is ____________
a) V/m2
b) MV/m2
c) MV/m
d) Vm
Answer: c
Explanation: Dielectric strength is the potential gradient required to cause a breakdown in the material. Potential gradient is the ratio of voltage and length, its unit is MV/m.


2. If the Voltage increases, what happens to dielectric strength?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero
Answer: a
Explanation: Dielectric strength is the potential gradient required to cause a breakdown in the material. Potential gradient is the ratio of voltage and length. Hence as potential increases, dielectric strength also increases.


3. If the potential difference in a material is 4MV and the thickness of the material is 2m, calculate the dielectric strength.
a) 2MV/m
b) 4MV/m
c) 6MV/m
d) 8MV/m
Answer: a
Explanation: Dielectric strength is the potential gradient required to cause a breakdown in the material. Potential gradient is the ratio of voltage and thickness.
Dielectric strength= V/t= 4/2= 2MV/m.


4. If the dielectric strength of a material is 4MV/m and its potential difference is 28MV, calculate the thickness of the material.
a) 4m
b) 7m
c) 5m
d) 11m
Answer: b
Explanation: Dielectric strength is the potential gradient required to cause a breakdown in the material. Potential gradient is the ratio of voltage and thickness.
V/dielectric strength= t= 28/4=7m.


5. If the thickness of the material increases, what happens to the dielectric strength?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero
Answer: b
Explanation: Dielectric strength is the potential gradient required to cause a breakdown in the material. Potential gradient is the ratio of voltage and thickness. Hence as thickness increases, dielectric strength decreases.


6. The thickness of a material having dielectric strength 10MV/m is 5m, calculate the potential difference.
a) 2MV
b) 10MV
c) 50MV
d) 100MV
Answer: c
Explanation: Dielectric strength is the potential gradient required to cause a breakdown in the material. Potential gradient is the ratio of voltage and thickness.
V=t*dielectric strength= 5*10=50MV.


7. Which medium has the highest dielectric strength?
a) Water
b) Mica
c) Air
d) Glass
Answer: c
Explanation: The better material is to prevent electrical conductivity, higher the dielectric strength. And the air is the best insulator so it has high dielectric strength.

Leakage and Conduction Currents in Capacitors

1. Leakage in capacitors is primarily caused by _________
a) Transistors
b) Resistors
c) Inductors
d) DC motors
Answer: a
Explanation: Leakage is primarily caused due to electronic devices, such as transistors, connected to the capacitors. Transistors conduct a small amount of current even when they are turned off, hence they are responsible for leakage current.


2. What is the conduction current when a capacitor is fully charged?
a) Infinity
b) Zero
c) 100A
d) 1000A
Answer: b
Explanation: When a capacitor is fully charged, there is no conduction of electrons from one plate of the capacitor to another, hence there is no conduction current and conduction current is equal to zero.


3. The flow of electrons in dielectric is due to _________
a) Conduction
b) Potential difference
c) Breakdown
d) Resistance
Answer: c
Explanation: There is, under normal circumstance, no flow of electrons in a dielectric since a dielectric is basically an insulator. Hence, there is a flow of electrons in a dielectric only at breakdown voltage.


4. The flow of electrons which does not pass through the battery is known as ________
a) Displacement current
b) Leakage current
c) Either displacement or leakage current
d) Neither displacement nor leakage current
Answer: a
Explanation: Displacement current is the flow of electrons from the positive plate of the capacitor to the negative plate of the capacitor, not through the battery. Hence the type of current which flows without passing through the battery is displacement current.


5. The free electrons in practical dielectrics is due to _________
a) There are no free electrons
b) Conductors
c) Impurities
d) Both conductors and impurities
Answer: c
Explanation: Ideally, dielectrics are insulators and do not contain any free electrons. But no dielectric is a perfect dielectric, hence the free electrons are due to impurities present in each dielectric.


6. The current in conductors connecting the voltage source to the plates of a capacitor is ______
a) Conduction current
b) Leakage current
c) Charging current
d) Zero
Answer: c
Explanation: The current in conductors connecting the voltage source to the plates of a capacitor is the charging current and not the conduction or leakage current.


7. What is the type of current where the electrons actually move?
a) Displacement current
b) Conduction current
c) Both conduction and displacement current
d) Neither conduction nor displacement current
Answer: b
Explanation: Conduction current is the current caused by the actual flow of electrons and displacement current is the current where no charge carriers are involved.


8. What is the type of current caused due to variations in the field?
a) Displacement current
b) Conduction current
c) Both conduction and displacement current
d) Neither conduction nor displacement current
Answer: a
Explanation: Displacement current is the current where no charge carriers are involved. It is caused due to variations in the electric field.


9. Under normal conditions capacitors have _________
a) Displacement current
b) Conduction current
c) Both conduction and displacement current
d) Neither conduction nor displacement current
Answer: a
Explanation: Under normal conditions capacitors contain an insulating material called dielectric sandwiched between the plates of the capacitor. Since insulators can carry only an electric field but not moving carriers, therefore normally a capacitor has displacement current and not conduction current.


10. If a large amount of voltage is applied to a capacitor, what is the current that flows through it?
a) Displacement current
b) Conduction current
c) Both conduction and displacement current
d) Neither conduction nor displacement current
Answer: b
Explanation: When a large amount of voltage is applied between the plates of a capacitor, the dielectric between the plates does not behave as an insulator anymore and starts conducting and conduction currents flow through it.

Displacement Current in a Dielectric

1. The current in conductors connecting the voltage source to the plates of a capacitor is _______
a) Conduction current
b) Leakage current
c) Charging current
d) Displacement current
Answer: c
Explanation: The current in conductors connecting the voltage source to the plates of a capacitor is the charging current and not the conduction or leakage current.


2. Under normal conditions capacitors have _______
a) Displacement current
b) Conduction current
c) Both conduction and displacement current
d) Neither conduction nor displacement current
Answer: a
Explanation: Under normal conditions capacitors contain an insulating material called dielectric sandwiched between the plates of the capacitor. Since insulators can carry only an electric field but not moving carriers, therefore normally a capacitor has displacement current and not conduction current.


3. What is the unit for displacement current?
a) No unit
b) Ampere
c) Coulomb
d) Ampere/coulomb
Answer: b
Explanation: Displacement current is a type of current and hence it has the same unit as that of current that is ampere.


4. Displacement current depends on ___________
a) Moving charges
b) Change in time
c) Both moving charges and change in time
d) Neither moving charges nor change in time
Answer: b
Explanation: Displacement current is the current which arises due to variations in the field. Hence, it does not depend on the moving charges but it changes with time which causes variation in the field.


5. Magnetic fields between the parallel plates of a capacitor are due to?
a) Displacement current
b) Conduction current
c) Both conduction and displacement current
d) Neither conduction nor displacement current
Answer: a
Explanation: Displacement current is the current which arises due to variations in the field. Change in the field results in the formation of magnetic fields. Hence displacement currents lead to magnetic field between the plates of a capacitor.


6. The free electrons in practical dielectrics are due to ________
a) There are no free electrons
b) Conductors
c) Impurities
d) Displacement currents
Answer: c
Explanation: Ideally, dielectrics are insulators and do not contain any free electrons. But no dielectric is a perfect dielectric, hence the free electrons are due to impurities present in each dielectric.


7. The flow of electrons which does not pass through the battery is known as ____________
a) Conduction current
b) Leakage current
c) Charging current
d) Displacement current
Answer: a
Explanation: Conduction current is the flow of electrons from the positive plate of the capacitor to the negative plate of the capacitor, not through the battery. Hence the type of current which flows without passing through the battery is conduction current.

Types of Capacitor and Capacitance

1. Paper capacitor is a type of _________
a) Fixed capacitor
b) Variable capacitor
c) Either fixed or variable depending on its usage
d) Neither fixed nor variable
Answer: a
Explanation: Paper capacitors are fixed capacitors because, like fixed capacitors, its capacitance value remains constant. In paper capacitors, paper is used as the dielectric.


2. A capacitor using chemical reactions to store charge is _______
a) Paper capacitor
b) Ceramic capacitor
c) Polyester capacitor
d) Electrolyte capacitor
Answer: d
Explanation: Electrolyte capacitors use chemical processes like pyrolysis to store charge between its plates.


3. Which, among the following, is the odd one out?
a) Ceramic capacitor
b) Electrolyte capacitor
c) Tuning capacitor
d) Paper capacitor
Answer: c
Explanation: Ceramic capacitor, electrolyte capacitor and paper capacitor are fixed capacitors whereas tuning capacitors is a variable capacitor, hence it is the odd one out.


4. In a variable capacitor, capacitance can be varied by ______
a) Turning the rotatable plates in or out
b) Sliding the rotatable plates
c) Changing the plates
d) Changing the material of plates
Answer: a
Explanation: As the plates are rotated, the area of the plates between which the field exists, will vary. Capacitance depends on area, hence as area varies, capacitance also varies.


5. The simplest kind of capacitor is ________
a) Ceramic capacitor
b) Electrolyte capacitor
c) Tuning capacitor
d) Paper capacitor
Answer: d
Explanation: The paper capacitor consists of two strips of aluminium foil separated by sheets of waxed paper. This whole setup is rolled up into the form of a cylinder. Since the materials requires for its construction are easily available, it is the simplest kind of capacitor.


6. Capacitor preferred when there is high frequency in the circuits is _______
a) Electrolyte capacitor
b) Mica capacitor
c) Air capacitor
d) Glass capacitor
Answer: b
Explanation: Mica capacitors are preferred for high frequency circuits because they have low ohmic losses and less reactance.


7. The type of capacitors used in communication transmitters are?
a) Electrolyte capacitor
b) Variable capacitor
c) Air capacitor
d) Glass capacitor
Answer: b
Explanation: Variable capacitor is used to tune all the circuits to same frequency i.e. resonance frequency so they are used in communication transmitters.


8. Which capacitors relatively costly?
a) Electrolyte capacitor
b) Mica capacitor
c) Air capacitor
d) Glass capacitor
Answer: b
Explanation: Mica capacitors are relatively expensive because it consists either of alternate layers of mica and metal foil clamped tightly together, or of thin films of silver on the two sides of a mica sheet. Silver is an expensive metal, hence mica capacitors are expensive.


9. ____________ capacitors usually have a colour code to find its value.
a) Electrolyte capacitor
b) Variable capacitor
c) Polyester capacitor
d) Glass capacitor
Answer: c
Explanation: Polyester capacitors usually come with a colour code because they are very small and their values cannot be printed on its body.


10. ______________ capacitors have a high leakage voltage.
a) Electrolyte capacitor
b) Variable capacitor
c) Air capacitor
d) Polyester capacitor
Answer: d
Explanation: Polyester capacitors can operate at high voltages, that is, a few thousand volts and the leakage resistance is high, that is, usually 100 M.

Capacitance And Capacitors MCQs ( Electrical Engineering ) MCQs – Competitive Electrical Engineering MCQs