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

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

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

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).

Concept Used:

The given network of capacitors forms a Wheatstone bridge , where the equivalent capacitance can be determined using series and parallel combinations.

Given:

C₁ = C₄ = 1 μF

C₂ = C₃ = 2 μF

Applying series and parallel capacitance rules:

Calculation:

The capacitors C₂ and C₃ are in parallel , so their equivalent capacitance is:

C_eq = (C₂ × C₃) / (C₂ + C₃)

C_eq = (2 × 2) / (2 + 2)

C_eq = 4 / 4 = 1 μF

Now, this C_eq is in series with C₁ and C₄:

C_AB = C₁ + C_eq + C₄

C_AB = 1 + 1 + 1

C_AB = 3 μF

∴ The equivalent capacitance between A and B is 3 μF.

Concept Used:

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

C_eq = C₁ + C₂ + C₃ + C₄

Calculation:

Given:

Each capacitor has capacitance C = 8 μF.

⇒ C_eq = 8 + 8 + 8 + 8

⇒ C_eq = 32 μF

∴ The correct answer is 32 μF.

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

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 = C1 + C2  + C3 +...  Cn

Calculation: 

Here 2F is in parallel with 2F,

So equivalent capacitor is 4F,

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

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

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

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

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

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.

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.

KVL in Capacitor:

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

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,

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

Concept:

A 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 correct answer is option 2):(infinite)

Concept:

The reactance of the capacitor is given by

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

where

f is the frequency in Hz

C is the capacitance in farad

Calculation:

Given 

for a DC source f = 0

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

= \(1 \over 2 \pi \times 0 \times 3\)

= \(\infty\)

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 

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

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

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

X_C 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',

X_C 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.

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.

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:

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