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Two identical capacitors of 12 pF each are connected in series across a 50 V battery. Calculate the electrostatic energy stored in the combination. If these were connected in parallel across the same battery, find out the value of the energy stored in this combination. |
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Answer» Solution :Here `C_1 = C_2 = 12 PF = 12 xx 10^(-12) F and V = 50 V`. When the two CAPACITORS are connected in series, the equivalent capacitance `C_s = (C_1C_2)/(C_1 +C_2) = ((12 xx 10^(-12))xx(12 xx 10^(-12)))/((12 xx 10^(-12) + 12 xx 10^(-12)))=6 xx 10^(-12) F` and energy stored in the series 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` When the same capacitors are connected in parallel, the equivalent capacitance `C_p =C_1 + C_2 = 12 xx 10^(-12) + 12 xx 10^(-12) = 24 xx 10^(12)F` `:.` Energy stored in parallel 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` |
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