Example

Question:

A network of four \(10\,\mu\text{F}\) capacitors is connected to a 500 V supply, as shown in Fig. 2.29. (a) Determine the equivalent capacitance of the network and the charge on each capacitor. (b) Find the charge on the plate with higher potential, equal and opposite to the charge on the plate with lower potential.

Solution:

(a)
In the given network, \(C_1, C_2, C_3\) are connected in series.
The effective capacitance \(C'\) of these three is \[ \frac{1}{C'} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \] For \(C_1=C_2=C_3=10\,\mu\text{F}\), \(C' = \frac{10}{3}\,\mu\text{F}\).
The network has \(C'\) and \(C_4\) in parallel, so \[ C = C' + C_4 = \left(\frac{10}{3} + 10\right)\,\mu\text{F} = 13.3\,\mu\text{F} \] (b)
Charge on each of the series capacitors \(Q\) and on \(C_4\), \(Q'\):
Let the charge on \(C_4\) be \(Q'\). The potential difference across AB is \(\frac{Q}{C_1}\), across BC is \(\frac{Q}{C_2}\), and CD is \(\frac{Q}{C_3}\).
\[ \frac{Q}{C_1} + \frac{Q}{C_2} + \frac{Q}{C_3} = 500\,\text{V}, \qquad Q'/C_4 = 500\,\text{V} \] So, \[ Q = 500\,\text{V} \times \frac{10}{3}\,\mu\text{F} = 1.7 \times 10^{-3}\,\text{C} \] \[ Q' = 500\,\text{V} \times 10\,\mu\text{F} = 5.0 \times 10^{-3}\,\text{C} \]