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Two identical capacitors of 12 pF each are connected in series across a battery of 50 V . How much electrostatic energy is stored in the combination ? If these were connected in parallel across the same battery , how much energy will be stored in the combination now ? Also find the charge drawn from the battery in each case. |
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Answer» Solution :Here `C_1 = C_2 = 12 pF` and battery voltage V = 50 V . When the capacitors are CONNECTED in series , the equivalent capacitance `C_s` is given as : `(1)/(C_(s)) = (1)/(C_(1)) + (1)/(C_(2)) = (1)/(12) + (1)/(12) = (1)/(6) implies C_(s) = 6 pF = 6 xx 10^(-12) F` `therefore` ENERGY STORED in the combination `U_(s) = (1)/(2) C_(s) V^(2) = (1)/(2) xx (6 xx 10^(-12)) xx (50)^(2) = 7.5 xx 10^(-9) J = 7.5 NJ` and charge drawn from the battery `Q_(s) = C_(s) V = 6 xx 10^(-12) xx 50 = 300 xx 10^(-12) C = 0.3 nC` If the capacitors are connected in parallel across the same battery , then the equivalent capacitance `C_(p) = C_(1)+ C_(2) = 12 + 12 = 24 pF = 24 xx 10^(-12) F` `therefore` Energy stored in the combination `U_(p) = (1)/(2) C_p V^(2) = (1)/(2) xx (24 xx 10^(-12)) xx (50)^(2) = 30 xx 10^(-9) J = 30 nJ` and charge drawn from the battery `Q_(p) = C_(p) V = (24 xx 10^(-12)) xx 50 = 1.2 xx 10^(-9) C = 1.2n C ` |
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