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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. |
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Answer» Solution :(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 ` |
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