Saturday, August 25, 2007

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).

You can find more posts on electrostatics by clicking on the label ELECTROSTATICS below this post or on the left side of this page.

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