Combination of Capacitors: Overview, Questions, Preparation

Q: Q. Three capacitors each of capacitance 9 pF are connected in series. (a) What is the total capacitance of the combination? (b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?

A Capacitance of each of the three capacitors, C = 9 pF The equivalent capacitance when the capacitors are connected in series is given by = + = = = Hence, = 3 pF Supply voltage, V = 120 V Potential difference ( ) across each capacitor is given by = = = 40 V

Q: Q. Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel. (a) What is the total capacitance of the combination? (b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.

A. Let the three capacitors be = 2 pF, = 3 pF and = 4 pF The equivalent capacitance, is given by = 2 + 3 + 4 = 9 pF When the combination is connected to a 100 V supply, the voltage in all three capacitors = 100 V charge in each capacitor is given by the relation, q = VC Hence for , = 100 = 200 pC = 2 C, for , = 100 = 300 pC = 3 C, for , = 100 = 400 pC = 4 C

Updated on Nov 17, 2023 11:26 IST

By Dhanya B

In NCERT Class 12 Physics, students learn about the combination of capacitors, which involves connecting multiple capacitors in different ways to achieve specific capacitance values or equivalent capacitance for a circuit. There are two common ways to combine capacitors: in series and in parallel.

Table of content

Combination of Capacitors in Series
Combination of Capacitors in Parallel
Applications of Series Combination and Parallel Combination
FAQs on Combination of Capacitors

Combination of Capacitors in Series

When capacitors are connected in series, they share the same charge. The equivalent capacitance (C_eq) for capacitors in series is less than the capacitance of individual capacitors. To find the equivalent capacitance of capacitors in series, use the following formula:

1 / C_eq = 1 / C₁ + 1 / C₂ + 1 / C₃ + ...

"C_eq" is the equivalent capacitance of the series combination.

"C₁," "C₂," "C₃," etc., are the individual capacitances of the capacitors in series.

It's essential to note that the inverse relationship in this formula makes the equivalent capacitance smaller than the smallest capacitance in the series.

Combination of Capacitors in Parallel

When capacitors are connected in parallel, they share the same voltage. The equivalent capacitance (C_eq) for capacitors in parallel is the sum of the individual capacitances. To find the equivalent capacitance of capacitors in parallel, use the following formula:

C_eq = C₁ + C₂ + C₃ + ...

"C_eq" is the equivalent capacitance of the parallel combination.

"C₁," "C₂," "C₃," etc., are the individual capacitances of the capacitors in parallel.

In this case, the equivalent capacitance is always greater than the largest individual capacitance in the parallel combination.

Applications of Series Combination and Parallel Combination

Series Combination

Series combinations are often used to create a lower overall capacitance value when required in electronic circuits.

They are also used for capacitors with different dielectric materials to achieve specific combinations.

Parallel Combination

Parallel combinations are used to increase the total capacitance when higher capacitance values are needed.

They are commonly found in applications like energy storage capacitors and power supply filter circuits.

Example:

Let's consider a practical example: You have three capacitors with values C₁ = 4 μF, C₂ = 3 μF, and C₃ = 6 μF.

If you connect them in series, you can calculate the equivalent capacitance as:

1 / C_eq = (1 / 4) μF + (1 / 3) μF + (1 / 6) μF

1 / Ceq = (6/12 + 4/12 + 2/12) μF

1 / C_eq = 12/12 μF

1 / C_eq = 1 μF

C_eq = 1 μF

If you connect them in parallel, the equivalent capacitance is simply the sum of the individual capacitances:

C_eq = 4 μF + 3 μF + 6 μF = 13 μF

So, when connected in series, the equivalent capacitance is 1 μF, and when connected in parallel, it's 13 μF. This demonstrates how the arrangement of capacitors affects the equivalent capacitance in a circuit.

FAQs on Combination of Capacitors

Q. Three capacitors each of capacitance 9 pF are connected in series. (a) What is the total capacitance of the combination? (b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?

Capacitance of each of the three capacitors, C = 9 pF

The equivalent capacitance when the capacitors are connected in series is given by

= + = = =

Hence, = 3 pF

Supply voltage, V = 120 V

Potential difference ( ) across each capacitor is given by = = = 40 V

Q. Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel. (a) What is the total capacitance of the combination? (b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.

Let the three capacitors be

= 2 pF, = 3 pF and = 4 pF

The equivalent capacitance, is given by

= 2 + 3 + 4 = 9 pF

When the combination is connected to a 100 V supply, the voltage in all three capacitors = 100 V

charge in each capacitor is given by the relation, q = VC

Hence for , = 100 = 200 pC = 2 C,

for , = 100 = 300 pC = 3 C,

for , = 100 = 400 pC = 4 C

Q. In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10–3 m2 and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?

A. Area of each plate, A = 6

Distance between plates, d = 3 mm = 3 m

Supply voltage, V = 100 V

Capacitance C of the parallel plate is given by C =

In case of air, dielectric constant k = 1 and

= permittivity of free space = 8.854

Hence C = F = 17.708 F = 17.71 pF

Charge on each plate of the capacitor is given by q = VC = 100 C

= 1.771 C

Therefore, capacitance of the capacitor is 17.71 pF and charge on each plate is 1.77 C

Q. Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates, (a) while the voltage supply remained connected. (b) after the supply was disconnected.

A. Dielectric constant of the mica sheet, k =6

While the voltage supply remained connected :

V = 100 V

Initial capacitance, C = 17.708 F

New capacitance, = kC = 6 17.708 = 106.25 F

= 106.25 pF

New charge, = = 106.25 C = 10.62 C

If the supply voltage is removed, then there will be constant amount of charge in the plates.

Charge, q = CV = 17.708 C = 1.7708 C

Potential across plates, = = = 16.66 V

Q. Obtain the equivalent capacitance of the network in Fig. 2.33. For a 300 V supply, determine the charge and voltage across each capacitor.

A. Let C’ be the equivalent capacitance for capacitors and connected in series.

Hence, = + . So C’ = 100 pF

Capacitors C’ and are in parallel, if the equivalent capacitance be C”, then

C” = C’ + = 100 + 100 = 200 pF

Now C” and are connected in series. If the total equivalent capacitance of the circuit be , then = + = + , = pF = F

Let V” be the potential difference across C” and be the potential difference across

Then, V” + = 300 V

Now, charge on is given by = = = 2 C

= = 200. So V” = 100 V

So potential difference across C’ and is V” = 100 V

Charge on is given by = 100 C = 1 C

and are having same capacitance and have a potential difference of 100V together. Since and are in series, the potential difference is given by = = 50V

Hence the charge on is given by = 200 = C and the charge on is given by = 200 = C

Therefore, the equivalent capacitance of the circuit is pF and charge and voltage at all capacitance is given as

For = C, = 100 V

For = C, = 50 V

For = 2 C, = 200 V