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

The correct answer is: 0.5 A

Key Points

• Maximum Current in AC Circuit:
  • The maximum current in a series RLC circuit is given by: I_max = V / R
  • Here, V = 6 V and I_max = 0.6 A
  • So, resistance R = V / I_max = 6 / 0.6 = 10 ohms
• Current through Coil with DC Cell:
  • Total resistance = resistance of coil + internal resistance of the cell
  • Given: resistance of coil = 10 ohms, internal resistance = 2 ohms
  • Total resistance = 10 + 2 = 12 ohms
  • Voltage from the cell = 6 V
  • So, current I = V / R_total = 6 / 12 = 0.5 A

Additional Information

• RLC Circuit in AC:
  • In AC, inductors and capacitors create reactance (opposition to current).
  • At resonance, inductive and capacitive reactances cancel out.
  • Only resistance (R) limits the current in that case.
• Ohm's Law:
  • Ohm's Law: I = V / R
  • This formula is used to find current when voltage and resistance are known.
  • It applies to both AC and DC circuits.

For a 1,000 V power supply, an electrolytic capacitor would be the most suitable choice. 

Concept:

• Electrolytic Capacitors:
  • These capacitors are designed for high voltage and high capacitance applications.
  • They have a high capacitance-to-volume ratio, making them efficient for filtering large amounts of ripple in high-voltage power supplies.
  • They are polarized, meaning they have a positive and negative terminal, which is important for DC power supplies
• • Air Capacitors: These are typically used in radio frequency (RF) applications where precise tuning is required, and they generally have low capacitance values. They are not suitable for high-voltage power supply filtering.
  • Mica Capacitors: These capacitors are known for their stability and precision, but they are typically used in high-frequency applications and are not ideal for high-voltage power supply filtering due to their lower capacitance values.
  Paper Capacitors: While they can handle moderate voltages, they are generally less efficient in terms of capacitance per volume compared to electrolytic capacitors.

Concept:

Energy stored in a charged capacitor is given by

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

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

After the insertion of the slab, the capacitance will be

C' = KC

Thus, the final energy is given by,

\(\;{U'} = \frac{1}{2}\frac{{{Q^2}}}{{C'}} = \frac{1}{2}\frac{{{Q^2}}}{{KC}}\)

\(U' = \frac{1}{K}U\)

So, energy dissipated in the process will be equal to work done on the slab, i.e.,

ΔU = U – U'

Calculation:

Given,

Capacitance of the capacitor C = 6 pF = 12 × 10-12 F

Potential difference V = 20V

The dielectric constant of the porcelain slab, K = 3.6

Energy stored in a charged capacitor is given by

\(U = \frac{1}{2}\frac{{{Q^2}}}{C}\)  ----(1)

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

Here C = 6 pF = 6 × 10-12 F and V = 20

\(\Rightarrow U = \frac{1}{2} \times 6 \times {10^{ - 12}} \times {20^2}\)

U = 12 × 10-10 J     ----(2)

After the insertion of the slab, the capacitance will be

C' = KC

Thus, the final energy is given by,

\(U' = \frac{1}{2}\frac{{{Q^2}}}{{C'}} = \frac{1}{2}\frac{{{Q^2}}}{{KC}}\)

\(U' = \frac{1}{K}U = \frac{1}{{3.6}} \times 12 \times {10^{ - 10}}J\)     ----(3)

[∵ given, K = 3.6 ]

So, energy dissipated in the process will be equal to work done on the slab, i.e.,

\({\rm{\Delta }}U = U - U' = \left( {12 \times {{10}^{ - 10}}} \right) - \left( {\frac{1}{{3.6}} \times 12 \times {{10}^{ - 10}}} \right)J\)

\({\rm{\Delta }}U = \left( {1 - \frac{1}{{3.6}}} \right) \times 12 \times {10^{ - 10}}J\)

\({\rm{\Delta }}U = \frac{{2.6}}{{3.6}} \times 12 \times {10^{ - 10}}J\)

ΔU = 8.6667 × 10-10 J

ΔU = 866.67 pJ

Concept:

If two capacitors (C₁,C₂) are in series then their equivalent capacitance(Ceq)

1/Ce_q =1/C₁ + 1/C₂

Reactance offered by a capacitor is given by;

X_C = 1/ωC = 1/2πfC Ω 

Current through the capacitor

I_C = V/X_C

Calculation:

Given;

C1 = 15 μf

C₂ = 30 μf

f = 50 Hz

V= 200 V

Then;

C_eq = (15× 30)/(15+30) = 10 μf

XC = 1/ωC = 1/2πfC = 1/(100π × 10 × 10^-6) = 1000/π Ω 

Therefore;

IC = V/XC = 200/(1000/π ) = (π/5) A

The correct answer is option 2):(Static Capacitors)

Concept:

The static capacitors provide a leading current that neutralizes (totally or approximately) the lagging inductive component of load current (i.e. leading component neutralizes or eliminates the lagging component of load current) thus power factor of the load circuit is improved

• Static capacitors are connected in parallel with those devices which work on low power factor
• These capacitors are installed in the vicinity of large inductive loads e.g. induction motors and transformers etc. and improve the load circuit power factor to improve the system or device efficiency

Additional Information

• A voltage multiplier is an electrical circuit that converts AC electrical power from a lower voltage to a higher DC voltage.
• Synchronous Condensers are used to improve stability and to maintain voltages within desired limits under changing load conditions and contingency situations.
• Phase Advancers are used to improve the power factor of induction motors. The low power factor of an induction motor is because its stator winding draws an exciting current that lags the supply voltage by 90°.

Concept:

If two capacitors (C₁,C₂) are in series then their equivalent capacitance(Ceq)

1/Ce_q =1/C₁ + 1/C₂

Reactance offered by a capacitor is given by;

X_C = 1/ωC = 1/2πfC Ω 

Current through the capacitor

I_C = V/X_C

Calculation:

Given;

C1 = 15 μf

C₂ = 30 μf

f = 50 Hz

V= 200 V

Then;

C_eq = (15× 30)/(15+30) = 10 μf

XC = 1/ωC = 1/2πfC = 1/(100π × 10 × 10^-6) = 1000/π Ω 

Therefore;

IC = V/XC = 200/(1000/π ) = (π/5) A

Concept:

Energy stored by the capacitor is

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

The relation between charge, potential difference & capacitance is

Q = C V

Q = Charge across the capacitor in Coulomb

V = Potential difference or voltage across the capacitor in Volts

C = Capacitance of the capacitor in Farad

Calculation:

Let capacitance across 3 capacitors are C₁,C₂ & C₃ respectively

C₁ : C₂ : C₃ = 1 : 2 : 3 (Given)

∴ C₂ = 2 C₁ & C₃ = 3 C₁

Let voltage across 3 capacitors are V₁,V₂ & V₃ respectively

V₁ : V₂ : V₃ = 3 : 2 : 1 (Given)

∴ V₂ = 2 V₃ & V₁ = 3 V₃

Now energy stored by capacitor C₁ is

\({E_{C1}} = \frac{1}{2}{C_1}V_1^2 = \frac{1}{2}{C_1}{\left( {3{V_3}} \right)^2}\)

\({E_{C1}} = \frac{9}{2}{C_1}V_3^2\)

Now energy stored by capacitor C₂ is

\({E_{C2}} = \frac{1}{2}{C_2}V_2^2 = \frac{1}{2}(2{C_1}){\left( {2{V_3}} \right)^2}\)

\({E_{C2}} = 4\;{C_1}V_3^2\)

Now energy stored by capacitor C₃ is

\({E_{C3}} = \frac{1}{2}{C_3}V_3^2 = \frac{1}{2}(3{C_1}){\left( {{V_3}} \right)^2}\)

\({E_{C3}} = \frac{3}{2}{C_1}V_3^2\)

Now the ratio of energy stored in the capacitors is

\({E_{C1}}\;:{E_{C2}}\;:{E_{C3}} = \;\frac{9}{2}:4\;:\frac{3}{2}\)

\({E_{C1}}\;:{E_{C2}}\;:{E_{C3}} = 9\;:8\;:3\)

The correct answer is option 2):  (6)

Concept:

When two capacitors are connected in parallel the equivalent capacitance is the sum of individual capacitance.

C_eq = C₁ + C₂ 

when two capacitors connected in series

 \(\frac{1}{Ceq} = \frac{1}{C_1} + \frac{1}{C_2}\)

Calculation:

Where 5 numbers of 10 μF capacitors connected in series, we get an equivalent resistance of:

\(\frac{1}{Ceq} = \frac{1}{10} + \frac{1}{10} + \frac{1}{10}+ \frac{1}{10}+ \frac{1}{10}\)

C_eq = 2 μF

Ceq = 2 μF connected in parallel with 10 μF capacitor. The equivalent resistance becomes:

C_total = 10 + 2 

= 12  μF

Testing of the capacitor:

While testing the capacitor using a multimeter, it measures the resistance offered by the capacitor.

• If the capacitor under test is short, the multi-meter needle will go to zero position right from the beginning.
• If the capacitor under test is open, the multi-meter needle will not move.
• If the capacitor under test has leakage then the needle will first deflect to zero, and then slowly move towards infinity and will settle at a point before infinity.

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: 

Current flows through a capacitor is given by

\(i = C\frac{{dV}}{{dt}}\)

Where C is the capacitance and V is the voltage across the capacitor.

\( V = \;\frac{1}{C}\mathop \smallint \nolimits_0^t i\;dt\)

Calculation:

Given:

i = 1mA

dV = 10 V

dt = 5 sec

\(∴ C=i\frac{dt}{dV}\)

\(C = 1mA\times\frac{5}{10}\)

C = 0.50 mF

Types of capacitors

1.) Electrolytic capacitor:

• An electrolytic capacitor is a polarized capacitor whose anode or positive plate is made of a metal that forms an insulating oxide layer through anodization.
• This oxide layer acts as the dielectric of the capacitor.
• Due to their very thin dielectric oxide layer and enlarged anode surface, electrolytic capacitors have a much higher capacitance-voltage (CV) product per unit volume.

2.) Dielectric capacitor:

• In this capacitor, a dielectric material is used to separate the conductive plates of the capacitor.
• The polarization of the dielectric by the applied electric field increases the capacitor's surface charge for the given electric field strength.

3.) Ceramic capacitor:

• A ceramic capacitor is a fixed-value capacitor where the ceramic material acts as the dielectric.
• It is constructed of two or more alternating layers of ceramic and a metal layer acting as electrodes.

4.) Film type capacitor:

• A film capacitor is defined as a capacitor that employs a thin plastic film as a dielectric. 
• These capacitors are not polarized, which is appropriate for AC signal and power applications.

Concept:

• In a purely resistive circuit, voltage and current are in the same phase i.e. angle between current and voltage is 0°.
•  In a purely inductive circuit, current lags behind voltage by an angle 90° i.e. angle between current and voltage is - 90°.
• In a purely capacitive circuit, current leads the voltage by an angle 90° i.e. angle between current and voltage is + 90°.

Application:

The voltage applied to a pure capacitor = V sin(4t)

Now current through the pure capacitor is

\(i = A\;sin\left( {4t + \frac{\pi }{2}} \right)\)

Where A is the peak value of current

Concept:

The charge stored by a capacitor with an applied voltage V is given by:

Q = C × V

C = Capacitance of the capacitor

V = Voltage applied across the capacitor

Q = Charge stored

Calculation:

Given Q = 0.5 C

V = 25 V

Putting on the respective values, we can write:

0.5 C = C × 25

\(C=\frac{0.5}{25}F\)

C = 0.02 F

The correct answer is option 2):(Static Capacitors)

Concept:

The static capacitors provide a leading current that neutralizes (totally or approximately) the lagging inductive component of load current (i.e. leading component neutralizes or eliminates the lagging component of load current) thus power factor of the load circuit is improved

• Static capacitors are connected in parallel with those devices which work on low power factor
• These capacitors are installed in the vicinity of large inductive loads e.g. induction motors and transformers etc. and improve the load circuit power factor to improve the system or device efficiency

Additional Information

• A voltage multiplier is an electrical circuit that converts AC electrical power from a lower voltage to a higher DC voltage.
• Synchronous Condensers are used to improve stability and to maintain voltages within desired limits under changing load conditions and contingency situations.
• Phase Advancers are used to improve the power factor of induction motors. The low power factor of an induction motor is because its stator winding draws an exciting current that lags the supply voltage by 90°.