Consider the three capacitors in series connected across the applied voltage V as shown in the Fig. 1. Suppose this pushes charge Q on C

_{1}then the opposite plate of C_{1}must have the same charge. This charge which is negative must have been obtained from the connecting leads by the charge separation which means that the charge on the upper plate of C_{2}is also Q. In short, all the three capacitors have the same charge Q.
If an equivalent capacitor stores the same charge, when applied with the same voltage, then it is obvious that,

_{1}, V

_{2}and V

_{3}if Q is known.

**Note**: For all the capacitors in series, the charge on all of them is always same, but the voltage across them is different.

__1. Voltage Distribution in Two Capacitors in Series__

Consider two capacitors C

_{1}and C_{2}connected in series. The voltage across them is say, V volts. This is shown in Fig. 2.Where V

_{1}is voltage across C

_{1}and V

_{2}is voltage across C

_{2}

Now , V = V

_{1}+ V

_{2}

From the expression of Q, we can write,

The result is exactly identical to the current distribution in two parallel resistances.

Note : Notice that the smallest capacitor has the largest of the three voltage across it and that C

_{eq}is lesser than any of the capacitors in the series string.
## 0 التعليقات:

## إرسال تعليق