Capacitors in Parallel and in Series MCQ Quiz - Objective Question with Answer for Capacitors in Parallel and in Series - Download Free PDF

Calculation:

C_eq = (ε₀ A) / (t₁/K₁ + t₂/K₂ + t₃/K₃)

Here C₀ = ε₀ A / d,   t₁ = 3d / 8,   t₂ = d / 2,   t₃ = d / 8

K₁ = K₁,   K₂ = K₁ / 1.25,   K₃ = 1

Given C_eq = 2C₀

⇒ 2C₀ = ε₀ A / ( (3d / 8K₁) + (d × 1.25 / 2K₁) + (d / 8) )

⇒ 2ε₀ A / d = ε₀ A / ( (3d / 8K₁) + (d / 2K₁) + (d / 8) )

⇒ 2 = 1 / ( (3 / 8K₁) + (5 / 8K₁) + (1 / 8) )

⇒ K₁ = 8 / 3 = 2.66

CONCEPT:

The capacitance of the capacitor is written as;

\(C_o = \frac{\epsilon _o A}{d}\)

where d is the distance and A is the area.

CALCULATION:

As we know,

\(C_o = \frac{\epsilon _o A}{d}\)

When \(C_1\) and \(C_2\) are in series.

\(\frac{1}{C'}=\frac{1}{C_1}+\frac{1}{C_2}\)

⇒ \(\frac{1}{C'}=\frac{3d}{ {4K\epsilon _o A}}+\frac{d}{4 {\epsilon _o A}}\)

⇒ \(\frac{1}{C'}=\frac{3d+K}{ {4K\epsilon _o A}}\)

⇒ \(\rm C'=\frac{{4K}}{K+3}C_0\)

Hence, option 1) is the correct answer.

Concept:

Capacitors in Series:

When two or more capacitors are connected end to end and have the same electric charge on each is called a series combination of capacitors.

Equivalent capacitance (Ceq) in series combination:

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

The charge on a capacitor is given by:

Charge (Q) = CV

Where C is capacitance and V is the potential difference.

Calculation:

Given:

C₁ = 2μF and C₂ = a 4μF

Equivalent capacitance (Ceq) in series combination:

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

\(\frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{4}=\frac{2+1}{4}=\frac{3}{4}\)

\({{{C_{eq}}}} = \frac{4}{{{3}}} \)

We know that charge is constant

Q = C_eqV

\(Q =\frac{4}{3}V\)

Let V₁, and V₂ be the potential difference across 2μF and 4μF respectively.

Q = CV₁ 

\(\frac{4V}{3}=2\times V_1\)

\(V_1=\frac{2V}{3}\)

Similarly;

Q = CV2 

\(\frac{4V}{3}=4\times V_1\)

\(V_2=\frac{V}{3}\)

\(V_1=2V_2\)

Thus 2μF capacitor has twice the potential difference as 4μF.

Concept:

Energy Stored in a Capacitor:

The energy U stored in a capacitor is given by the formula:

U = 1/2 × Q × V, where:

Q = Charge on the capacitor (Coulombs)

V = Voltage across the capacitor (Volts)

The graph given in the question represents the relationship between charge Q and voltage V on the capacitor. The area of the triangle OAB represents the energy stored in the capacitor, as the energy stored is the product of charge and voltage divided by 2.

Calculation:

The area of triangle OAB is given by:

Area = 1/2 × base × height

Here, base = Q (charge) and height = V (voltage). Thus, the area is equivalent to the energy stored in the capacitor.

∴ The area of triangle OAB represents the energy stored in the capacitor, which corresponds to Option 4.

Ans.(4)

Sol.

\(C_{0}=\frac{\in_{0} A}{d} \ldots \ldots \ldots (1)\)

\(C=\frac{7}{6} C_{0} \ldots \ldots \ldots (2)\)

\(C=\frac{\epsilon_{0} A}{d-t\left(1-\frac{1}{K}\right)}=\frac{\epsilon_{0} A / d}{1-\frac{t}{d}\left(1-\frac{1}{K}\right)}\)

\(\frac{C_{0}}{1-\frac{2}{3}\left(1-\frac{1}{K}\right)}=\frac{3 K C_{0}}{2+K} \ldots \ldots (3)\)

Dividing (1) and (3)

⇒ \(\frac{c}{C_{0}}=\frac{3 K}{K+2}=\frac{7}{6}\)

⇒ \(K=\frac{14}{11}\)

CONCEPT:

• The capacitance of a capacitor (C): The capacity of a capacitor to store the electric charge is called capacitance.
  • The capacitance of a conductor is the ratio of charge (Q) to it by a rise in its potential (V).

​C = Q/V

• 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 and V = Electric potential

Combination of capacitors:

• Parallel combination: When two or more capacitors are connected in such a way that their ends are connected at the same two points and have an equal potential difference for all capacitor is called a parallel combination of the capacitor.

• Equivalent capacitance (C_eq) for parallel combination:

C_eq = C₁ + C₂ + C₃

Where C₁ is the capacitance of the first capacitor, C₂ is the capacitance of the second capacitor and C₃ is the capacitance of the third capacitor

• Series combination: When two or more capacitors are connected end to end and have the same electric charge on each is called the series combination of the capacitor.

​

• Equivalent capacitance (Ceq) in series combination:

\(\frac{1}{{{C_{eq}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \frac{1}{{{C_3}}}\)

EXPLANATION:

• The energy stored in the capacitor is 0.5 C V². So statement 1 is correct.
• As the charge stored in the capacitor is given by Q = C V. Therefore the energy stored in the capacitor is 0.5 QV. So statement 2 is correct.
• The charge stored in the capacitor is given by Q = C V. So statement 3 is correct.
• When n capacitors are connected in series combination then equivalent capacitance is 

CONCEPT:

• The device that stores electrical energy in an electric field is called a capacitor.
  • The capacity of a capacitor to store electric charge is called capacitance.
• When two or more capacitors are connected in such a way that their ends are connected at the same two points and have an equal potential difference for all capacitor is called the parallel combination of a capacitor.
  • Equivalent capacitance (C_eq) for parallel combination:

Ceq = C1 + C₂

• When two or more capacitors are connected end to end and have the same electric charge on each is called a series combination of the capacitor.
  • Equivalent capacitance (Ceq) in series combination:

\(\frac{1}{{{C}_{eq}}}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}\)

Where C1 and C2 are two capacitors in the circuit.

EXPLANATION:

• From the above, it is clear that when three capacitors are connected in parallel, then the value equivalent capacitance (Ceq)  is increased.
• The energy stored in the capacitor is

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

Where U = energy stored in the capacitor, C = capacitance of the capacitor, and V = Electric potential difference

• From the above equation, it is clear that the energy stored in the capacitor is directly proportional to the capacitance of the capacitor when the voltage remains the same.
• Hence, three capacitors should be connected in parallel to get high energy on the same potential. Therefore option 3 is correct.

CONCEPT:

Capacitor:

• The capacitor is a device in which electrical energy can be stored.
  • In a capacitor two conducting plates are connected parallel to each other and carrying charges of equal magnitudes and opposite sign and separated by an insulating medium.
  • The space between the two plates can either be a vacuum or an electric insulator such as glass, paper, air, or semi-conductor called a dielectric.​

​1. Capacitors in series

• When two or more capacitors are connected one after another such that the same charge gets generated on all of them, then it is called capacitors in series.
• The net capacitance/equivalent capacitance (C) of capacitors in series is given by,

\(\Rightarrow\frac{1}{C} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}}+...+ \frac{1}{{{C_n}}}\)

2. Capacitors in parallel

• When the plates of two or more capacitors are connected at the same two points and the potential difference across them is equal, then it is called capacitors in parallel.
• The net capacitance/equivalent capacitance (C) of capacitors in parallel is given by,

\(\Rightarrow C = C_1+ C_2+...+ C_n\)

CALCULATION:

Given C1 = C2 = C3 = C4 = C5 = C 

• When they are connected in series the equivalent capacitance is given as,

\(\Rightarrow\frac{1}{C_{S}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}}+ \frac{1}{{{C_3}}}+ \frac{1}{{{C_4}}}+ \frac{1}{{{C_5}}}=\frac{5}{C}\)

\(\Rightarrow C_{S} = \frac{C}{5}\)     -----(1)

• When they are connected in parallel the equivalent capacitance is given as,

\(\Rightarrow C_P = C_1+ C_2+ C_3 +C_4+C_5 = 5C\)    ----- (2)

By dividing equation 1 and equation 2,

\(\Rightarrow \frac{C_{S}}{C_{P}}=\frac{C}{5\times 5C}=\frac{1}{25}\)

• Hence, option 3 is correct.

CONCEPT:

• The capacitance of a capacitor (C): The capacity of a capacitor to store the electric charge is called capacitance.
  • The capacitance of a conductor is the ratio of charge (Q) to it by a rise in its potential (V).

​C = Q/V

• The unit of capacitance is the farad, (symbol F ).
• Series combination: When two or more capacitors are connected end to end and have the same electric charge on each is called the series combination of the capacitor.

​

• Equivalent capacitance (Ceq) in series combination:

\(\frac{1}{{{C_{eq}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \frac{1}{{{C_3}}}\)

CALCULATION:

​Given that:

C₁ = 3 μF, C2 = 2 μF, and C3 = 6 μF

Potential of the battery (V) = 10 V

They are in series combination:

So \(\frac{1}{{{C_{eq}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \frac{1}{{{C_3}}}\)

\(\frac{1}{{{C_{eq}}}} = \frac{1}{{{3}}} + \frac{1}{{{2}}} + \frac{1}{{{6}}} =\frac{2 + 3 +1}{6}=\frac{6}{6}= 1\)

C_eq = 1 μF

Charge (Q) = C V = 1 × 10 = 10 μC

Since they are in series combination, so the charge on each capacitor will be the same.

So charge on 3 μF = 10 μC

Hence option 1 is correct.

EXTRA POINTS:

• Parallel combination: When two or more capacitors are connected in such a way that their ends are connected at the same two points and have an equal potential difference for all capacitor is called a parallel combination of the capacitor.

• Equivalent capacitance (Ceq) for parallel combination:

Ceq = C1 + C2 + C3

Where C1 is the capacitance of the first capacitor, C2 is the capacitance of the second capacitor and C3 is the capacitance of the third capacitor

CONCEPT:

Capacitor:

• The capacitor is a device in which electrical energy can be stored.
  • In a capacitor two conducting plates are connected parallel to each other and carrying charges of equal magnitudes and opposite sign and separated by an insulating medium.
  • The space between the two plates can either be a vacuum or an electric insulator such as glass, paper, air, or semi-conductor called a dielectric.

​Capacitance

• The charge on the capacitor (Q) is directly proportional to the potential difference (V) between the plates,

​⇒ Q ∝ V

⇒ Q =  CV

Where C = capacitance

Series combination of capacitors

• When capacitors are connected in series, the magnitude of charge Q on each capacitor is the same. The potential difference across C1 and C2 is different.
• The equivalent capacitance is given as,

\(⇒ \frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}\)

Where V = total potential difference

CALCULATION:

Given C₁ = 6 μF, C₂ = 4 μF and V = 120 V

• In a series combination of the capacitor,

\(⇒ \frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}\)

\(⇒ \frac{1}{C}=\frac{1}{6}+\frac{1}{4}\)

⇒ C = 2.4 μF     -----(1)

• We know that for the series arrangement of the capacitor,

⇒ Q =  CV = C₁V₁ = C₂V₂    -----(2)

• Hence, option 1 is correct.

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.

• And in the Series circuit, the reciprocal of the equivalent capacitance is the algebraic sum of all the reciprocal of the capacitance.

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

CALCULATION:

Given that two equal capacitors of capacitance 2 F are connected in series.

Ceq = 2 F

Let the equal capacitors are C'

So their effective capacitance when connected in series

1/Ceq = 1/C1 + 1/C2

1/Ceq = 1/C' + 1/C'

1/2C = 2/C'

C' = 2 × 2 F = 4 F

Each capacitors have capacity of 4 F.

So the correct answer is option 1.

Additional Information 

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.
• And 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)

CALCULATION:

Given that three capacitors are  C1 = C and C2 = nC

• The effective capacitance when connected in parallel

⇒ Ceq = C1 + C2 

⇒ Ceq = C + nC = C(1 + n)

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.

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

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

CALCULATION:

Given that three equal capacitors of capacitance C are connected in series.

Ceq = 5 μF

So their effective capacitance when connected in series

\(\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\)

\(\frac{1}{C_{eq}}=\frac{1}{C}+\frac{1}{C}+\frac{1}{C}\)

\( {1 \over C_{eq}} = {3 \over C}\)

C_eq = \(\frac{C}{3}\)

⇒ 5 =  \(\frac{C}{3}\)

⇒ C = 15 μF

Hence the correct answer is option 1.

Mistake PointsIt is not the individual capacitance that is given to us. It is the equivalent capacitance is given to us, i.e. C_eq = 5 μF. 

Concept: 

Energy stored in capacitor = \(\frac{1}{2}CV^{2} = \frac{1}{2}QV\)   ---- (1)

Electrostatic energy stored in capacitor = \(\frac{1}{2}(C_{1} + C_{2})V^{2}\)    ---- (2)

The common potential is \(V_{c} = \frac{C_{1}V_{1} + C_{2}V_{2}}{C_{1}+C_{2}}\)   ---- (3)

Calculation: 

Given: 

Voltage of battery = 100 V , Capacitance of capacitor = C = 900 pF

Initially energy stored in capacitor = \(\frac{1}{2}CV^{2} = \frac{1}{2}QV\) = (1/2)× 900× 10^-12 × (100)² = 4.5× 10^-6 J

When the capacitor is disconnected and is connected to another capacitor of 900 pF which can be seen as:

Charge on the system becomes Q' = 2Q

From equation (3) we get: 

The common potential is \(V_{c} = \frac{C_{1}V_{1} + C_{2}V_{2}}{C_{1}+C_{2}}\)

⇒ \(V_{c} = \frac{C× 100 + C× 0}{C+C}\) = 50v

The electrostatic energy stored in capacitor = \(\frac{1}{2}(C_{1} + C_{2})V^{2}\) = \(\frac{1}{2}900× 10^{-12}× 2×(50)^{2}\) = 2.25× 10^-6 J

Hence option 4) is correct.

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)

CALCULATION:

Given that: 

C₁ = C₂ = 12 F

They are in series combination, 

1/Ceq = 1/C1 + 1/C2

1/Ceq = 1/12 + 1/12 = 2/12 = 1/6

So C_eq = 6 F

Hence option 3 is correct.