Two Questions (MCQ) on Capacitors

Here are two multiple choice questions which will be interesting to you:

(1) Six identical capacitors C₁, C₂, C₃, C₄, C₅ and C₆ each having capacitance C are connected as shown. The equivalent capacitance between the points A and B is

(a) C (b) 2C (c) 2.5C (d) 3C (e) 3.5C

This may appear to be a difficult question for some of you. In questions of this type try to identify equipotential points. Circuit elements connected between equipotential points can be ignored. In most cases the circuit will then become simple.

In the present case, the junction between C₁ and C₆ is at the same potential as that at the junction between C₂ and C₅. [This follows since the capacitors are identical. Generally, the points will be equipotential points if the balance condition for Wheatstone bridge is satisfied: C₁/C₆ = C₂/C₅]. The capacitor C₃ connected between the equipotential points can therefore be ignored.

Now we have three parallel connections across the points A and B:

(i) Series combination of C₁ and C₆ giving a capacitance C/2.

(ii) Series combination of C₂ and C₅ giving a capacitance C/2.

(iii) Capacitor C₄ of value C.

The parallel combined value of the abve three connections is (C/2) + (C/2) + C = 2C.

(2) In a circuit, capacitors C₁ and C₂ (in series) are connected between two points A and B (Fig). If the potentials at A and B are V₁ and V₂ respectively, what is the potential at the junction point P between C₁ and C₂?

(a) C₁(V₁+V₂)/(C₁+C₂)

(b) (V₁ + V₂)/2

(c) V₁ – V₂

(d) (C₁V₂ + C₂V₁)/ (C₁ + C₂)

(e) (C₁V₁ + C₂V₂)/ (C₁ + C₂)

Since the capacitors are in series, they carry the same charge. If the potential at the junction point P is ‘V’, the charges on the capacitors C₁ and C₂ are respectively C₁(V – V₁) and C₂(V₂ – V).

Therefore, we have

C₁(V – V₁) = C₂(V₂ – V), from which

V = (C₁V₁ + C₂V₂)/ (C₁ + C₂).

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