#### (a) Capacitors connected in parallel

Figure shows three capacitors, C_{1}, C_{2} and C_{3}, connected in parallel with a supply voltage V applied across the arrangement.

When the charging current I reaches point A it divides, some flowing into C_{1}, some flowing into C_{2 }and some into C_{3}. Hence the total charge Q_{T }( = I × t) is divided between the three capacitors.

The capacitors each store a charge and these are shown as Q_{1}, Q_{2} and Q_{3} respectively. Hence

Q_{T} = Q_{1} + Q_{2} + Q_{3}

But Q_{T} = CV, Q_{1} = C_{1}V, Q_{2} = C_{2}V and Q_{3} = C_{3}V. Therefore CV = C_{1}V + C_{2}V + C_{3}V where C is the total equivalent circuit capacitance, i.e.

C = C_{1} + C_{2} + C_{3}

It follows that for n parallel-connected capacitors,

C = C_{1} + C_{2} + C_{3} ······ + C_{n}

i.e. the equivalent capacitance of a group of parallel connected capacitors is the sum of the capacitances of the individual capacitors. (Note that this formula is similar to that used for resistors connected in series).

#### (b) Capacitors connected in series

Figure shows three capacitors, C_{1}, C_{2} and C_{3}, connected in series across a supply voltage V. Let the p.d. across the individual capacitors be V_{1}, V_{2} and V_{3} respectively as shown.

Let the charge on plate ‘a’ of capacitor C_{1 }be +Q coulombs. This induces an equal but opposite charge of −Q coulombs on plate ‘b’. The conductor between plates ‘b’ and ‘c’ is electrically isolated from the rest of the circuit so that an equal but opposite charge of +Q coulombs must appear on plate ‘c’, which, in turn, induces an equal and opposite charge of −Q coulombs on plate ‘d’, and so on.

Hence when capacitors are connected in series the charge on each is the same. In a series circuit:

V =V_{1} +V_{2} +V_{3}

Since V=\frac{Q}{C} then \frac{Q}{C}=\frac{Q}{C_1}+\frac{Q}{C_2}+\frac{Q}{C_3}

where C is the total equivalent circuit capacitance, i.e.

\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}It follows that for n series-connected capacitors:

\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\cdots+\frac{1}{C_{\mathrm{n}}}i.e. for series-connected capacitor, the reciprocal of the equivalent capacitance is equal to the sum of the reciprocals of the individual capacitances. (Note that this formula is similar to that used for resistors connected in parallel).

For the special case of two capacitors in series:

\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}=\frac{C_2+C_1}{C_1 C_2}Hence

C=\frac{C_1 C_2}{C_1+C_2} \quad\left(\text { i.e. } \frac{\text { product }}{\text { sum }}\right)Read More Topics |

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