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InterviewSolution
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Two capacitors of unknown capacitances C_1 and C_2 are connected first in series and then in parallel across a battery of 100 V . If the energy stored in the two combinations is 0.045 J and 0.25 J respectively , determine the value of C_1 and C_2 . Also calculate the charge on each capacitor in parallel combination . |
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Answer» <P> Solution :When CAPACITORS , `C_1` and `C_2` are joined in series then the equivalent capacitance `C_(s) = (C_(1) C_2)/(C_(1) + C_(2))` and in parallel arrangement the equivalent capacitance `C_(p) = ( C_(1) + C_(2))` . As voltage V = 100 V , energy stored by series combination `U_(s) = 0.045` J and energy stored by parallel combination `U_(p) = 0.25 J` , hence we have`U_(s)= (1)/(2) C_(s) V^(2) = (1)/(2) XX ((C_(1) C_(2))/(C_(1) + C_(2)))xx (100)^(2) = 0.045"" ..... (i)` and `U_(p) = (1)/(2) C_(p) V^(2) = (1)/(2) xx (C_(1) + C_(2)) xx (100)^(2) = 0.25 "" ..... (ii)` On solving EQUATIONS (i) and (ii) , we get : `C_(1) = 38.2 mu F` and `C_(2) = 11.8 mu F` In parallel combination voltage ACROSS each capacitor is same at V = 100 V `therefore` Charge on 1st capacitor `Q_(1)= C_(1) V = 38.2 xx 100 mu C = 3.82 mC` and charge on 2nd capacitor `Q_(2) = C_(2) V = 11.8 xx 100 mu C = 1.18 m C ` |
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