Multiple Choice Questions on Unusual Combinations of Capacitors

"It is unwise to be too sure of one's own wisdom. It is healthy to be reminded that the strongest might weaken and the wisest might err."

– Mahatma Gandhi

Today we will consider a few questions on unconventional combinations of capacitors which occasionally find place in entrance examination question papers conducted for admitting students to medical, engineering and other degree courses.

(1) Four parallel metal plates are arranged in air as shown, with the same separation ‘d’ between neighbouring plates. If the area (of one surface) of each plates is A, the capacitance between the terminals T₁ and T₂ is

(a) 4ε₀A/d

(b) ε₀A/d

(c) 2ε₀A/d

(d) 3ε₀A/d

(e) ε₀A/3d

Let us number the plates starting from the top as 1, 2, 3, and 4 respectively. The lower surface of plate No.1 and the upper surface of plate No.2 with air in between makes a capacitor of capacitance ε₀A/d.

The lower surface of plate No.2 and the upper surface of plate No.3 with air in between makes another capacitor of capacitance ε₀A/d.

The lower surface of plate No.3 and the upper surface of plate No.4 with air in between makes a third capacitor of capacitance ε₀A/d.

This arrangement therefore makes three capacitors in parallel. The capacitance value of each capacitor is ε₀A/d so that the parallel combined value which appears between the terminals T₁ and T₂ is 3ε₀A/d [Option (d)].

(2) If the connection of metal plates in the above question is modified as shown in the adjoining figure, the capacitance between terminals T₁ and T₂ is

(a) ε₀A/2d

(b) 3ε₀A/2d

(c) 2ε₀A/d

(d) 4ε₀A/d

(e) 2ε₀A/3d

In this case the capacitor at the centre (having capacitance ε₀A/d) appears directly across the terminals T₁ and T₂. The upper and lower capacitors (each of value ε₀A/d) are connected in series and this series combination having effective capacitance ε₀A/2d also appears between the terminals T₁ and T₂. Therefore, the effective capacitance between T₁ and T₂ is ε₀A/d + ε₀A/2d = 3ε₀A/2d [Option (b)].

(3) If a fifth plate is added to the combination shown in question No.1, as shown in the figure, what is the effective capacitance between T₁ and T₂?

(a) ε₀A/4d

(b) ε₀A/5d

(c) 5ε₀A/d

(d) 4ε₀A/5d

(e) 4ε₀A/d

Once you answer question No.1, this question becomes quite simple. You can easily see that there are four identical capacitors (each having capacitance ε₀A/d) connected in parallel. Therefore, the effective capacitance between T₁ and T₂ is 4ε₀A/d [Option (e)].

You can find a few more questions (with solution) in this section here.

Karnataka CET Multiple Choice Questions on Combination of Capacitors

The following question appeared in Karnataka Common Entrance Test (CET) 2010 question paper. Even though at the first reading it may appear somewhat difficult for you on seeing the circuit, a careful look at the circuit will assure you that it’s simple.

All capacitors used in the diagram are identical and each is of capacitance C. Then the effective capacitance between the points A and B is

(a) 1.5 C

(b) 6 C

(c) C

(d) 3 C

The first three capacitors are in parallel, giving an effective capacitance 3 C. The last three capacitors also are in parallel, giving an effective capacitance 3 C.. These two parallel combinations are connected in series. Therefore, the effective capacitance between the points A and B is 3 C/2 = 1.5 C.

The following question appeared in the Karnataka CET 2008 question paper:

How many 6 μF, 200 V condensers are needed to make a condenser of 18 μF, 600 V?

(1) 9

(2) 18

(3) 3

(4) 27

Since the voltage rating of each 6 μF capacitor is 200 V, you should connect 3 capacitors in series to obtain the required voltage rating of 600 V. But then the effective capacitance of the series combination of these three capacitors will be 2 μF (6/3 = 2). To obtain the required capacitance of 18 μF, you need to connect nine such series combinations in parallel. So the total number of capacitors required is 9×3 = 27.

The following question also appeared in the Karnataka CET 2008 question paper:

The total energy stored in the condenser system shown in the figure will be

(1) 2 μJ

(2) 4 μJ

(3) 8 μJ

(4) 16 μJ

The series combination of 6 μF and 3 μF gives an effective capacitance of (6×3)/(6+3) = 2 μF. Since this is in parallel with a 2 μF capacitor, the effective capacitance C across the 2 volt battery is 4 μF. The energy stored in the system of capacitors is ½ CV² = (½)×(4×10^–6) ×2² = 8×10^–6 J = 8 μJ.

You will find a few more multiple choice questions (with solution) in this section here.