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?
Answer (Detailed Solution Below)
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.
Answer (Detailed Solution Below)
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
Answer (Detailed Solution Below)
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.
⇒ 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 -
Answer (Detailed Solution Below)
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
Answer (Detailed Solution Below)
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?
Answer (Detailed Solution Below)
Capacitance Question 6 Detailed Solution
Download Solution PDFConcept:
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,
\(C_{AB} = \frac{(6\times2)}{8} = 1.5 F\)
_______ capacitors have the largest capacitance values per volume of element of any capacitor type.
Answer (Detailed Solution Below)
Capacitance Question 7 Detailed Solution
Download Solution PDFElectrolytic 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.
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.
Answer (Detailed Solution Below)
Capacitance Question 8 Detailed Solution
Download Solution PDFKVL 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,
\(V_{C_2}=V\times \frac{C_1}{C_1+C_2}=200\times \frac{5}{10+5}=66.66 V\)
What would be the energy required to charge a capacitor of 100 μF from 100 V to 500 V?
Answer (Detailed Solution Below)
Capacitance Question 9 Detailed Solution
Download Solution PDFConcept:
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 reactance of a 3 F capacitor when connected to a DC source as shown in the figure will be:
Answer (Detailed Solution Below)
Capacitance Question 10 Detailed Solution
Download Solution PDFThe 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
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 _________.
Answer (Detailed Solution Below)
Capacitance Question 11 Detailed Solution
Download Solution PDFConcept:
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 μFThe energy stored in the capacitor C1 under steady state condition is:
Answer (Detailed Solution Below)
Capacitance Question 12 Detailed Solution
Download Solution PDFThe 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
One of the main applications of a Capacitance is to
Answer (Detailed Solution Below)
Capacitance Question 13 Detailed Solution
Download Solution PDFA 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.
Answer (Detailed Solution Below)
Capacitance Question 14 Detailed Solution
Download Solution PDFConcept:
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:
Answer (Detailed Solution Below)
Capacitance Question 15 Detailed Solution
Download Solution PDFConcept
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:
\(V=5\times {10\over 10+40}=1\space V\)