In the circuit shown the capacitors are` C_1 = 15mu F , C_2 = 10 mu F and C_2 = 25 mu F . `Find

(i) The equivalent capacitance of the circuit
(ii) The charge on each capacitor and
(iii) The potential difference a across each capacitor.

Explanation

(i) ` ( ( C_1+C_2)C_3)/( (C_1+C_2)+C_3) = ( ( 25xx25)/( 25+25)) mu F = 12.5mu F `
` (##MOT_CON_JEE_PHY_C25_SLV_017_S01.png" width="80%">
(ii) Q = Total charge supplied by the cell = CV = ` (12.5xx 10 ) mu C = 125 mu C `
charge on
` C_1 = Q_1 = ((C_1)/((C_1+C_2)) Q= ( ( 15)/( 15+10 ) ) xx 125 = 75 muC `
Charge on
` C_2 = Q_2 = ( ( C_2 )/( C_1+C_2)) Q= ( (10 )/( 15+ 10 )) xx 125 = 50 mu C `
(iii) p.d. across ` C_1 = V_1 = (Q_1 )/(C_1 ) = ( 75)/( 15) = 5V `
p.d. across `C_2 =V_2 V-1 = 5V `
p.d. across ` C_2 = V_2 =(Q_3)/(C_3) = (125)/( 25) = 5V `