Capacitors in Parallel Questions

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Very Short Answers Questions [1 Mark Questions]

Ques. Calculate the overall capacitance when 3 capacitors of magnitude 2 F, 6 F, and 10 F are connected in parallel.

1. 20 F
2. 15 F
3. 18 F
4. 22 F

Ans. The correct answer is c. 18 F

Explanation: The capacitance of each capacitor gets added in parallel combination to give equivalent capacitance.

C_eq = 2 + 6 + 10 = 18 F

Ques. Two capacitors of capacitance 1 µF and 2 µF are connected in parallel across the voltage of 100 V. Calculate the charge on the capacitor of capacitance 2 µF.

1. 200 µC
2. 300 µC
3. 250 µC
4. 350 µC

Ans. The correct answer is a. 200 µC

Explanation: Since both capacitors are connected in parallel, therefore potential across each capacitor will be the same and it is equal to the applied voltage. Therefore charge on the capacitor of capacitance 2 µF will be

Q = 2 x 100 = 200 µC

Ques. The overall capacitance of capacitors connected in parallel will be always_____ the individual capacitance.

1. Equal to
2. Greater than
3. Lesser than
4. Cannot be determined

Ans. The correct answer is b. Greater than

Explanation: The overall capacitance for a parallel combination of capacitors will be the sum of the capacitance of individual capacitors. Therefore overall capacitance is always greater than the capacitance of individual capacitors.

Ques. How to arrange three capacitors in a circuit to get high energy on the same potential?

1. Two in series and one in parallel
2. All three in parallel
3. All three in series
4. Two in parallel in one parallel

Ans. The correct answer is b. All three in parallel

Explanation: The electrostatic energy stored in a combination of capacitors is directly proportional to the equivalent capacitance of the combination. Since the equivalent capacitance of the capacitors is greater if the capacitors are connected in parallel. Therefore to get high energy all three capacitors should be connected in parallel.

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Short Answers Questions [2 Marks Questions]

Ques. Four capacitors of capacitance 4 µF, 10 µF, 20 µF, and 56 µF are connected in parallel. What is the total capacitance of the combination?

Ans. Given

• C₁ = 4 µF
• C₂ = 10 µF
• C₃ = 20 µF
• C₄ = 56 µF

The formula for the equivalent capacitance for the capacitors connected in parallel is given by

C_eq = C₁ + C₂ + C₃ + C₄

⇒ C_eq = 4 + 10 + 20 + 56 = 90 µF

Ques. What is common potential? 

Ans. When two capacitors of different capacities are connected and charged to different potentials connected in parallel by a metallic wire, charge flows from a capacitor at the higher potential to the capacitor at a lower potential. This flow of charge continues till the potential of both capacitors becomes equal. This equal potential of both capacitors is known as Common potential.

Ques. Define capacitors in parallel. Write the formula for capacitors in parallel.

Ans. Two or more capacitors are said to be connected in parallel if one side of all the capacitors is connected to one terminal of a battery and the other side of all the capacitors is connected to the other terminal of the battery such that the potential difference across each capacitor is the same.

Let three capacitors C₁, C₂, and C₃ be connected in parallel then the equivalent capacitance of the combination is given by

C_eq = C₁ + C₂ + C₃

Ques. Define capacitors in series. Write the formula for capacitors in series.

Ans. Two or more capacitors are said to be connected in series if the left side of a capacitor is connected to the right side of another capacitor such that the magnitude of the charge on each capacitor is the same.

Let three capacitors C₁, C₂, and C₃ be connected in series then the equivalent capacitance of the combination is given by

1/C_eq = 1/C₁ + 1/C₂ + 1/C₃

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Long Answers Questions [3 Marks Questions]

Ques. In the arrangement shown, find the equivalent capacitance between A and B.

Ans. First, mark the different junctions (nodes) in the circuit. A node is a point where more than two wires are connected.

As we can clearly see, capacitors 10 µF and 20 µF are connected between the same points C and B. Hence they are in parallel combination.

The capacitors 10 µF and 20 µF can be replaced by a single capacitor of capacitance

C = 10 µF + 20 µF = 30 µF

Thus the circuit can be shown as

Now the capacitors 15 µF and 30 µF are connected in series. Therefore equivalent capacitance is given by

1/C_eq = 1/15 + 1/30 = 3/30

⇒ C_eq = 10 µF

Ques. Three capacitors of capacitance 30 µF, 50 µF, and 80 µF are connected in parallel to a 100 V battery. Calculate the energy stored in the capacitors.

Ans. Given

• C₁ = 30 µF
• C₂ = 50 µF
• C₃ = 80 µF
• V = 100 V

Since the capacitors are connected in parallel, therefore equivalent capacitance is given by

C_eq = C₁ + C₂ + C₃

⇒ C_eq = 30 µF + 50 µF + 80 µF = 160 µF = 160 x 10^-6 F

The energy stored in the series or parallel combination of capacitors is given by

E = C_eqV²/2

⇒ E = (160 x 10^-6 x 100²) / 2 = 0.8 J

Ques. Differentiate between parallel and series combinations of capacitors.

Ans. The differences between parallel and series combinations of capacitors are

Ques. For the following arrangement, find the equivalent capacitance between A and B.

Ans. The given circuit can be redrawn as

Here the capacitors 3 µF, 4 µF, and 5 µF are connected in series, therefore equivalent capacitance of the series combination is given by

1/C = 1/3 + 1/4 + 1/5 = 47/60

⇒ C = 60/47 µF

Now the capacitors 2 µF and 60/47 µF are connected in parallel, therefore the equivalent capacitance of the parallel combination is given by

C_eq = 2 + 60/47 = 154/47 µF = 3.28 µF

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Very Long Answers Questions [5 Marks Questions]

Ques. Derive the equation for energy stored in a parallel combination of capacitors.

Ans. Let V be the potential applied across the n capacitors connected in parallel.

The equivalent capacitance of capacitors in parallel is given by

C_eq = C₁ + C₂ + C₃ +...........+ C_n

The energy stored in a capacitor is given by

U = CV²/2

Therefore

• The energy stored in the capacitor C₁ is U₁ = C₁V²/2
• The energy stored in the capacitor C₂ is U₂ = C₂V²/2
• The energy stored in the capacitor C₃ is U₃ = C₃V²/2
• The energy stored in the capacitor C_n is U_n = C_nV²/2

The total energy stored in the parallel combination of capacitors is

⇒ U_Total = U₁ + U₂ + U₃ +...........+ U_n

⇒ U_Total = C₁V²/2 + C₂V²/2 + C₃V²/2 +.........+ C_nV²/2

⇒ U_Total = (C₁ + C₂ + C₃ +...........+ C_n)V²/2

⇒ U_Total = C_eqV²/2

Ques. What is the equivalent capacitance of a capacitor in parallel?

Ans. Let three capacitors C₁, C₂, and C₃ be connected in parallel. Let V be the potential difference applied across the combination of the capacitors, then the potential difference across each capacitor will be V.

Let q₁, q₂, and q₃ be the charge on capacitors C₁, C₂, and C₃ respectively.

The total charge on the combination of capacitors 

q = q₁ + q₂ + q₃ ….(i)

Also, we have

• q₁ = C₁V
• q₂ = C₂V
• q₃ = C₃V

Substituting the above values in equation (i), we get

q = C₁V + C₂V + C₃V

⇒ q = V(C₁ + C₂ + C₃)

⇒ q/V = (C₁ + C₂ + C₃) ….(ii)

Let C be the equivalent capacitance of the combination, then

q/V = C

Therefore, equation (ii) becomes

C = C₁ + C₂ + C₃

If n number of capacitors are connected in parallel then, the equivalent capacitance of the combination is given by

C_equivalent = C₁ + C₂ + C₃ +........+ C_n

Ques. Derive an expression for the loss of energy on sharing of charges between two capacitors.

Ans. When two capacitors of capacitance C₁ and C₂ are connected in parallel, no charge is lost but some energy is lost on sharing charges between two capacitors. This loss of energy appears as heat.

Let V₁ and V₂ be the potential across the capacitors, then electrostatic potential energy before sharing charges is given by

U_i = C₁V₁²/2 + C₂V₂²/2

Electrostatic energy after sharing charges is given by

U_f = C₁V₂/2 + C₂V₂/2 = (C₁ + C₂)V²/2

Where V is the common potential of the combination after sharing charges.

But the common potential is given by

V = (C₁V₁ + C₂V₂) / (C₁ + C₂)

U_f = \(\frac{(C_1+C_2)}{2} (\frac{C_1V_1 + C_2V_2}{C_1 + C_2})^2 = \frac{(C_1V_1 + C_2V_2)^2}{2(C_1 + C_2)}\)

On sharing charges, some energy is dissipated in the form of heat. Therefore loss of energy is given by

U_i – U_f = \(\frac{C_1V_1^2}{2}+ \frac{C_2V_2^2}{2} - \frac{(C_1V_1+C_2V_2)^2}{2(C_1+C_2)}\)

U_i – U_f = \(\frac{C_1C_2(V_1-V_2)^2}{2(C_1+C_2)}\)

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