Capacitance MCQ Quiz - Objective Question with Answer for Capacitance - Download Free PDF

Last updated on May 15, 2025

Latest Capacitance MCQ Objective Questions

Capacitance Question 1:

Determine the amount of charge stored on 5 μF capacitor when it is connected to a 12 V battery?

  1. 4.8 μC
  2. 30 μC
  3. 60 μC
  4. 2.4 μC

Answer (Detailed Solution Below)

Option 3 : 60 μC

Capacitance Question 1 Detailed Solution

Concept

The amount of charge stored in a capacitor is given by:

\(Q=CV\)

where, Q = Charge

C = Capacitance

V = Voltage

Calculation

Given, C = 5 μF

V = 12 volts

\(Q=5 \times 12=60\space \mu C\)

Capacitance Question 2:

In an ac circuit, the current leads the voltageby π/2. The circuit is _________.

Fill in the blank with the correct answer from the options given below.

  1. purely resistive
  2. should have circuit elements with resistance equal to reactance.
  3. purely inductive
  4. purely capacitive

Answer (Detailed Solution Below)

Option 4 : purely capacitive

Capacitance Question 2 Detailed Solution

Concept:

In an AC circuit, the phase difference between voltage and current indicates the type of circuit element. If the current leads the voltage by \(\frac{\pi}{2}\), it corresponds to a purely capacitive circuit.

Explanation: 

In a purely capacitive AC circuit, the voltage lags the current by \(\frac{\pi}{2}\) radians. This means that the current reaches its maximum value a quarter cycle before the voltage does. This phase relationship is characteristic of a capacitor in an AC circuit.

For a purely resistive circuit, the voltage and current are in phase (i.e., there is no phase difference). In a purely inductive circuit, the current lags the voltage by \(\frac{\pi}{2}\) radians. In circuits where resistance is equal to reactance, the phase difference would be neither \(\frac{\pi}{2}\) nor zero, but some intermediate value.

The correct option is (4).

Capacitance Question 3:

Four condensers each of capacitance 8 µF are joined as shown in the figure. The equivalent capacitance between the points A & B will be
qImage67911b95ad586e276f795a55

  1. 32 µF
  2. 2 µF
  3. 8 µF
  4. 16 µF

Answer (Detailed Solution Below)

Option 1 : 32 µF

Capacitance Question 3 Detailed Solution

Concept Used:

For capacitors connected in parallel, the total capacitance is given by:

Ceq = C1 + C2 + C3 + C4

Calculation:

Given:

Each capacitor has capacitance C = 8 μF.

3aei71k57m5dk-s

⇒ Ceq = 8 + 8 + 8 + 8

⇒ Ceq = 32 μF

∴ The correct answer is 32 μF.

Capacitance Question 4:

A capacitor carries a charge of 0.1 C at 5 V. Its capacitance is -

  1. 0.02 F
  2. 0.5 F
  3. 0.05 F
  4. 0.2 F

Answer (Detailed Solution Below)

Option 1 : 0.02 F

Capacitance Question 4 Detailed Solution

Concept

The capacitance of a capacitor is given by:

\(C={Q\over V}\)

where, C = Capacitance

Q = Charge

V = Voltage

Calculation

Given, Q = 0.1C

V = 5 volts

 

 

 

 

\(C={0.1\over 5}\)

C = 0.02 F

Capacitance Question 5:

Five capacitors each of value 1uF are connected as shown in the figure .The equivalent capacitance between A and B is

qImage67069baa06608ec6ea4a8acd5-5-2025 IMG-1217 Shubham Kumar Tiwari -8

  1. 1 μF
  2. 2 μF
  3. 5 μF
  4. 3 μF

Answer (Detailed Solution Below)

Option 1 : 1 μF

Capacitance Question 5 Detailed Solution

Given:

Five capacitors each of value 1μF

Connected in a specific arrangement

Concept Used:

Series and Parallel Capacitors:

In series, the reciprocal of the total capacitance is the sum of the reciprocals of the individual capacitances.

In parallel, the total capacitance is the sum of the individual capacitances.

For series: 1/Ceq = 1/C1 + 1/C2 + ...

For parallel: Ceq = C1 + C2 + ...

Calculation:

Assume the connection of capacitors given:

Two capacitors in series: 1/C1 = 1/1 + 1/1 ⇒ C1 = 0.5μF

Two capacitors in series: 1/C2 = 1/1 + 1/1 ⇒ C2 = 0.5μF

Now, C1 and C2 in parallel with a single 1μF capacitor:

Ceq = 0.5 + 0.5 + 1

∴ The equivalent capacitance is 2μF.

Top Capacitance MCQ Objective Questions

What will be the equivalent capacitance at the terminals A, B?

F29 Shubham B 19-4-2021 Swati D20

  1. 1.5 F
  2. 2.5 F
  3. 4 F
  4. 3 F

Answer (Detailed Solution Below)

Option 1 : 1.5 F

Capacitance Question 6 Detailed Solution

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Concept:

Capacitor: 

  • A capacitor is an electrical component with two terminals used to store charge in the form of an electrostatic field. 
  • It consists of two parallel plates each possessing equal and opposite charges, separated by a dielectric constant.
  • Capacitance is the ability of a capacitor to store charge in it. The capacitance C is related to the charge Q and voltage V across them as:

\(⇒ C =\frac{Q}{V} \)

Equivalent capacitance of capacitors:

Connected in series: When n capacitors C1, C2, C3, ... Cn are connected in series, the net capacitance (Cs) is given by:

\(⇒ \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2}+ \frac{1}{C_3} + ... \frac{1}{C_n}\)

Connected in parallel: When n capacitors C1, C2, C3, ... Cn are connected in parallel, the net capacitance (Cp) is given by:

⇒ Cp = C+ C2  + C+...  Cn

Calculation: 

Here 2F is in parallel with 2F,

F1 Nakshatra 20.5.21 Pallavi D6

So equivalent capacitor is 4F,

This 4F is again parallel with 4F, then equivalent is 8F

F1 Nakshatra 20.5.21 Pallavi D7

Again this 8F in series with 8F, then equivalent is 4F.

F1 Nakshatra 20.5.21 Pallavi D8

This 4F is parallel with 2F so equivalent is 6F,

F1 Nakshatra 20.5.21 Pallavi D9

This 6F is series to 2F, then equivalent capacitance across A and B is,

F1 Nakshatra 20.5.21 Pallavi D10

\(C_{AB} = \frac{(6\times2)}{8} = 1.5 F\) 

 _______ capacitors have the largest capacitance values per volume of element of any capacitor type.

  1. Mica
  2. Paper
  3. Ceramic
  4. Electrolytic

Answer (Detailed Solution Below)

Option 4 : Electrolytic

Capacitance Question 7 Detailed Solution

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Electrolytic Capacitor:

  • It’s a polarized capacitor whose anode or positive plate is made of metal
  • The solid, liquid, or gel electrolyte covers the surface of the oxide layer serving as a cathode or negative plate.

There are three families:

1. Aluminium electrolytic capacitors

2. Tantalum electrolytic capacitors

3. Niobium electrolytic capacitors

  • These are generally used when very large capacitance values are required.
  • Due to their very thin oxide layer and enlarged anode surface, electrolytic capacitors have a much higher capacitance-voltage product per unit volume than ceramic or film capacitors.

 

F1 Shubham.B 21-10-20 Savita D 19.

F1 Shubham.B 21-10-20 Savita D 20

Electrolytics are widely used capacitors due to their low cost and small size but there are 3 easy ways to destroy this

1. overvoltage

2. reversed polarity

3. over temperature

The most commonly used dielectric is “Aluminium oxide”.

The main disadvantage is it can’t be used on AC supplies.

If two capacitors having capacitance of 5 μF and 10 μF respectively are connected in series across a 200 V supply, find the potential difference across each capacitor.

  1. 50 V, 100 V
  2. 133.33 V, 66.66 V
  3. 200 V, 100 V
  4. 100 V, 200 V

Answer (Detailed Solution Below)

Option 2 : 133.33 V, 66.66 V

Capacitance Question 8 Detailed Solution

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KVL in Capacitor:

Consider two capacitors of capacitance C1 and C2 connected in series across supply having impedance Z1 and Z2 respectively as shown.

F1 Shubham Madhuri 11.05.2021 D17

Applying Voltage division rule to the circuit,

The voltage across C1 is given as,

\(V_{C1}=V\times \frac{Z_1}{Z_1+Z_2}\) .... (1)

The voltage across C2 is given as,

\(V_{C1}=V\times \frac{Z_2}{Z_1+Z_2}\) .... (2)

The impedance Z1 and Z2 can be written as,

\(Z_1=\frac{1}{\omega C_1},Z_2=\frac{1}{\omega C_2}\)

Put the value of Z1 and Z2 in equation (1) and (2),

\(V_{C1}=V\times \frac{\frac{1}{\omega C_1}}{\frac{1}{\omega C_1}+\frac{1}{\omega C_2}}\)

Hence, the voltage across C2 will be,

\(V_{C_1}=V\times \frac{C_2}{C_1+C_2}\)

Similarly, Voltage across C2 will be,

\(V_{C_2}=V\times \frac{C_1}{C_1+C_2}\)

Application:

Given,

C1 = 5 μF
C2 = 10 μF
V = 200 V

From the above concept, the voltage across C1 is given as,

\(V_{C_1}=V\times \frac{C_2}{C_1+C_2}=200\times \frac{10}{10+5}=133.33 V\)

And the voltage across C2 is given as,

\(V_{C_2}=V\times \frac{C_1}{C_1+C_2}=200\times \frac{5}{10+5}=66.66 V\)

What would be the energy required to charge a capacitor of 100 μF from 100 V to 500 V?

  1. 24 J
  2. 12 J
  3. 18 J
  4. 6 J

Answer (Detailed Solution Below)

Option 2 : 12 J

Capacitance Question 9 Detailed Solution

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Concept:

capacitor is a device used to store energy.

The process of charging up a capacitor involves the transferring of electric charges from one plate to another.

The work done in charging the capacitor is stored as its electrical potential energy.

The energy stored in the capacitor is

\(U = \;\frac{1}{2}\frac{{{Q^2}}}{C} = \frac{1}{2}C{V^2} = \frac{1}{2}QV\)

Where,

Q = charge stored on the capacitor

U = energy stored in the capacitor

C = capacitance of the capacitor

V = Electric potential difference

Calculation:

Capacitance (C) = 100 μF  = 100 × 10-6 F

Applied voltage V1= 100 V and V2 = 500 V

The energy stored in the capacitor is

\(\frac{1}{2} \times 100\times 10^{-6} \times \left( {V_{2\;}^2 - V_1^2} \right)\)

\(\frac{1}{2} \times 100\times 10^{-6} \times \left( {{{500}^2} - {{100}^2}} \right)\)

= 12 J

The reactance of a 3 F capacitor when connected to a DC source as shown in the figure will be:

F3 Vinanti Engineering 03.01.23 D9

  1. 10 Ω 
  2. infinite
  3. 0.1 Ω
  4. zero

Answer (Detailed Solution Below)

Option 2 : infinite

Capacitance Question 10 Detailed Solution

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The correct answer is option 2):(infinite)

Concept:

The reactance of the capacitor is given by

Xc = \(1 \over 2 \pi \times f \times C\)

where

f is the frequency in Hz

C is the capacitance in farad

Calculation:

Given 

F3 Vinanti Engineering 03.01.23 D9

for a DC source f = 0

Xc = \(1 \over 2 \pi \times f \times C\)

\(1 \over 2 \pi \times 0 \times 3\)

\(\infty\)

A capacitor charged to 200 V has 2000 μC of charge. The value of capacitance will be _________.

  1. 100 F
  2. 100 μF
  3. 10 μF
  4. 10 F

Answer (Detailed Solution Below)

Option 3 : 10 μF

Capacitance Question 11 Detailed Solution

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Concept:

For a voltage V applied across the capacitor, the charge stored by it is given by:

Q = C × V

C = Capacitance (depends upon the physical dimensions)

\(C=\frac{Q}{V}\)

Calculation:

Given that, Q = 2000 μC

V = 200 V

⇒ 2000 × 10-6 = C × 200

⇒ C = 10 μF

The energy stored in the capacitor C1 under steady state condition is:

F4 Vinanti Engineering 29.12.22 D10

  1. 350 J
  2. 100 J
  3. 25 J
  4. 250 J

Answer (Detailed Solution Below)

Option 4 : 250 J

Capacitance Question 12 Detailed Solution

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The correct answer is option 4):(250 J)

Concept:

The energy stored in the capacitor is

\(\Rightarrow U = \frac{1}{2}C{V^2}\)

Calculation:

Given the circuit is 

F4 Vinanti Engineering 29.12.22 D10

The voltage across the capacitor is  5V

The energy stored in the capacitor is 

U =  \(1\over 2\) CV2

\(1\over 2\) × 20 × 52

= 250 J

One of the main applications of a Capacitance is to

  1. block ac and pass dc
  2. block both dc and ac
  3. block dc and pass ac
  4. pass both dc and ac

Answer (Detailed Solution Below)

Option 3 : block dc and pass ac

Capacitance Question 13 Detailed Solution

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A capacitor passes AC signals and blocks DC signals.

This can be understood with the help of the reactance formula.

The capacitive reactance is given by:

\(X_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}\)

f = frequency of the voltage/current applied across the capacitor.

C = Capacitance value

For DC:

A DC signal is a zero frequency signal, i.e. f = 0 Hz

XC for f = 0 will be:

\(X_C=\frac{1}{2\pi (0)C}= \infty \)

For AC:

AC signal is a signal having a particular frequency 'f',

XC for any frequency f is given by:

\(X_C=\frac{1}{2\pi fC}= Finite\)

Observation:

  • For DC we conclude that a capacitor provides infinite resistance to the flow of current. Hence no current will flow as current in a capacitive circuit is given by:

          \(I=\frac{V}{X_C}\)

  • For AC we observe that the resistance offered by the capacitor is finite. Hence a finite flow of current is possible in a capacitive circuit with AC input.

A constant current of 2 mA charges a 20 μF capacitor for 2s. Which of the following is true for the charging of the capacitor.

  1. Capacitor voltages increases linearly from 0V to 200V
  2. Capacitor voltage increases exponentially from 0V to 200V
  3. Capacitor voltage increases linearly from 0V to 100V
  4. Capacitor voltage increases exponentially from 0V to 100V

Answer (Detailed Solution Below)

Option 1 : Capacitor voltages increases linearly from 0V to 200V

Capacitance Question 14 Detailed Solution

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Concept:

The current across a capacitor is defined as:

\(i = C.\frac{{d{V_c}}}{{dt}}\)

Rearranging the above, we can write:

\(\frac{{d{V_c}}}{{dt}} = \frac{i}{C}\)

Integrating both sides we get:

\({V_c}\left( t \right) = \frac{1}{C}\mathop \smallint \nolimits_0^t i.dt\)

Calculation:

Given i = 2 mA (Constant)

C = 20 μF

The capacitor voltage will be given by:

\({V_c}\left( t \right) = \frac{1}{20~\mu F}\mathop \smallint \nolimits_0^t 2~mA~dt\)

\({V_c}\left( t \right) = \frac{1}{20~\mu F}.2~mA(t)|_0^t\)

\({V_c}\left( t \right) =100t\)

This indicates a linear variation of the voltage across the capacitor.

The voltage after t = 2 sec will be:

\({V_c}\left( 2 \right) =100\times 2=200~V\)

So, the capacitor voltage increases linearly from 0 to 200 V in 2 sec.

For the circuit shown below, the voltage across 10F and 40 F capacitors are: 

F1 Savita ENG 26-8-24 D1 V2

  1. 1 V and 4 V, respectively
  2. 4 V and 1 V, respectively
  3. 400 V and 1600 V, respectively
  4. 10 V and 40 V, respectively 

Answer (Detailed Solution Below)

Option 2 : 4 V and 1 V, respectively

Capacitance Question 15 Detailed Solution

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Concept

When two capacitances are connected in parallel, their equivalent capacitance is given by:

\(C=C_1+C_2 \)

When two capacitances are connected in series, their equivalent capacitance is given by:

\(C={C_1C_2\over C_1+C_2}\)

Calculation

In the given circuit, 5F and 3F are connected in parallel.

\(C=5+3=8F\)

10F and 40F are connected in series.

\(C={10\times 40\over 10+40}=8F\)

When two equal capacitances are connected in series, the voltage gets equally divided between them.

∴ The combination of C3 and C4 will have a 5V voltage.

The voltage across C3 is given by:

\(V=5\times {40\over 10+40}=4\space V\)

The voltage across C4  is given by:

\(V=5\times {10\over 10+40}=1\space V\)

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