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Two capacitors of capacitance of 6 muF and 12 muF are connected in series with a battery. The voltage across the 6 muF capacitor is 2 V.Compute the total battery voltage. |
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Answer» Solution :In series ARRANGEMENT of CAPACITORS `V_1/V_2 = C_2/C_1` `rArr V_2 = (C_1 V_1)/(C_2) = (6muF xx 2V)/(12muF) = 1V` `:.` TOTAL voltage across the capacitors COMBINATION ` = V_1 + V_2 = 2 + 1 =3V` `:.` Battery voltage = 3 V 
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