Capacitance MCQ Quiz in தமிழ் - Objective Question with Answer for Capacitance - இலவச PDF ஐப் பதிவிறக்கவும்

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:

• Let C be the capacitance of each capacitor.

• When 4 identical capacitors are connected in series, the equivalent capacitance, Cs \(⇒ \frac{1}{C_s} = \frac{1}{C} + \frac{1}{C}+ \frac{1}{C} + \frac{1}{C} =\frac{4}{C}\)

Thus, the equivalent capacitance in series, \(C_s=\frac{4}{C}\)

• When 4 same capacitors are connected in parallel, the equivalent capacitance, Cp = C1 + C2  + C3 + C4 = 4C

Thus, the equivalent capacitance in parallel, Cp = 4C

∴ Ratio of the capacitances connected in parallel to series arrangement

= \(\frac{C_p}{C_s} = \frac{4C}{(C/4)} = 16\)

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 = 0.12 C

V = 10 V

Putting on the respective values, we can write:

0.12 C = C × 10

\(C=\frac{0.12}{10}F\)

C = 0.012 F

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:

Given - Capacitance (C) = 5.0 μF = 5 × 10^-6 F and voltage applied (V) = 100 V

The energy stored in the capacitor is

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

\( \Rightarrow U = \frac{1}{2} \times 5 \times {10^{ - 6}} \times {\left( {100} \right)^2}\)

\( \Rightarrow U = \frac{1}{2} \times 5 \times {10^{ - 6}} \times {10^4} = 2.5 \times {10^{ - 2}}J\)

Concept:

The formula for capacitance (C) is given as

\(C = \;\frac{{A{ϵ_0}{ϵ_r}}}{d}\)

Where, A = Area of parallel plate

d = Distance between the plate

ϵ₀ = Permittivity of free space = 8.85 × 10^-12 F/m

ϵ_r =  Permittivity of the medium

Energy \(E = \frac{1}{2}C{V^2}\)

Calculation:

Given, A = 10 cm2 = 10 × 10-4 m2

d = 0.8854 cm = 0.8854 × 10-2

ϵ_r = 5

\(C = \frac{{10 × {{10}^{ - 4}} × 8.85 × {{10}^{ - 12}×{5}}}}{{{{0.8854}×{10}^{ - 2}}}} = 5 × {10^{ - 12}}F\)

Capacitance (C) =  5 pF

Concept

The current flowing through a capacitor having stored charge(Q) is:

\(I={Q\over t}\)

where, I = Current

Q = Charge

t = Time 

Calculation

Given, Q = 70 μC

I = 3.5 A

\(3.5={70\times 10^{-6}\over t}\)

t = 20 μsec

CONCEPT:

Capacitor: 

• A capacitor is a two-terminal energy storage device that stores energy in the form of a static electric field during the positive half cycle and gives away during the negative half cycle of supply.

Parallel plate capacitor:

• When two parallel plates are connected across a battery, the plates are charged and an electric field is established between them, and this setup is known as the parallel plate capacitor.

​EXPLANATION:

The capacitance of a parallel plate capacitor: 

• The capacitance is the capacity of the capacitor to store charge in it.
• Two conductors are separated by an insulator (dielectric) and when an electric field is applied, electrical energy is stored in it as a charge.
• In capacitors, the metal plates are used to connect a voltage source between the dielectric medium. Hence, the electric charge is produced due to the presence of a dielectric medium and it is stored in that dielectric medium itself.

Concept:

When capacitors are connected in parallel, the total capacitance is the sum of the individual capacitance values, i.e.

\({C_{eq}}(\text{parallel}) = {C_1} + {C_2} + \ldots + {C_n}\)

Also, when capacitors are connected in series, the total capacitance is given by:

\(\frac{1}{{{C_{eq}(\text{series})}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \ldots + \frac{1}{{{C_n}}}\)

Analysis:

For the given circuit, C₁ and C₂ are in series. The equivalent capacitance will be:

\(\frac{1}{{{C_{eq}(\text{series})}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}}\)

\(C_{eq1}=\frac{C_1\times C_2}{C_1+C_2}=\frac{1}{2}=0.5 ~μ F\)

This 1/2 μF is now connected parallel to C₃. The equivalent capacitance will be:

C_eq2 = 0.5 + 1 μF

C_eq2 = 1.5 μF

The circuit is now redrawn as:

The required equivalent capacitance will be the series combination of the above two capacitances, i.e.

\(C_{eq}=\frac{1.5\times 1}{1.5+1}=\frac{1.5}{2.5}=\frac{3}{5}\)

C_eq = 0.6 μF

CONCEPT

Capacitance: The capacitance tells that for a given voltage how much charge the device can store.

Q = CV

where Q is the charge in the capacitor V is the voltage across the capacitor and C is the capacitance of it.​

In the Parallel circuit, the equivalent capacitance is the algebraic sum of all the capacitance.

In the Series circuit, the reciprocal of the equivalent capacitance is the algebraic sum of all the reciprocal of the capacitance.

Ceq = C1 + C2 + C3 +......  (In parallel)

1/Ceq = 1/C1 + 1/C2 + 1/C3 +...... (in series)

Calculations:

C₁ = 15 μA, C₂ = 5 μA, V₁ = 400 V, V₂ = 200 V

q₁ = C₁ × V₁

= 15 × 10^-6 × 400 = 6 mC

q₂ = C₂ × V₂

= 5 × 10^-6 × 200

= 1 mC

q_eq = q1 + q₂ = 7 mC

C_eq = C₁ + C₂ = 15 + 5 = 20 μ F

qeq = Ceq V_eq

7 × 10^-3 = 20 × 10^-6 × Veq

Veq = 350 V

• A ceramic capacitor is a capacitor which uses a ceramic material as the dielectric.
• There are two classes of ceramic capacitors available today: class 1 and class 2.
• Class 1 ceramic capacitors are used where high stability and low losses are required.
• They are very accurate and the capacitance value is stable in regard to applied voltage, temperature and frequency.
• Ceramic capacitors can also be used as a general purpose capacitor, since they are not polarized and are available in a large variety of capacitances, voltage ratings and sizes.

Capacitive reactance:

Capacitive reactance is given by:

\({X_C} = \frac{1}{{ω C}}\)

With ω = 2πf

\({X_C} = \frac{1}{{2π fC}}\)

\({X_C} \propto \frac{1}{{ f}}\)

The capacitive reactance is inversely proportional to the frequency.

Inductive reactance:

Inductive reactance (XL) is given by:

XL = ωL

ω = frequency in radian/sec which can be written as:

ω = 2πf

XL = ωL = 2πfL

f  = frequency in Hz

L = value of inductor in Henry

From the above, we conclude that:

XL ∝ f (for a constant L)

The inductive reactance is directly proportional to frequency.