A network of four capacitors, each of capacitance `15muF`,is connected across a battery of 100 V as shown in the figure. Find the net capacitance and the charge on the capacitor `C_4`.

Explanation

Here `C_1 = C_2 = C_3 = C_4= 15 muF and V = 100V`
In the network shown,capacitors `C_1, C_2 andC_3` are joined in series and their equivalent capacitance C. is given by :
`1/(C.) = 1/C_1 + 1/C_2 + 1/C_3 = 1/15+ 1/15+ 1/15 = 1/5 rArr C. = 5 muF`.
Now, C. and `C_4` are connected in parallel, hence the net capacitance of the network :
`C = C. + C_4 = (5 + 15) mu F = 20 muF`
Moreover, the charge on the capacitor `C_4`,
`Q_4 = C_4 V = (15 muF) xx 100V =1500 muC = 1.5 mC`