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We have two capacitors C_1 and C_2of capacitance 12 muF and 6 muF respectively. As shown in the given circuit arrangement the two capacitors are joined to a power supply of 6 volts. Initially the switch S_1 is closed but switch S_2 is open. After some time, the switch S_1 is opened and simultaneously switch S_2 is closed. Now answer the following questions : The fractional loss in electrostatic energy of the arrangement after closing the switch S_2 is |
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Answer» `1/4` `u_i= 1/2 C_1 V_1^2 = 1/2 XX (12 mu F) xx (6 V)^2 = 126muJ` and final value of electrostatic energy of the arrangement `u_f = 1/2 (C_1 + C_2) V^2 = 1/2 (12 muF+ 6 mu F) xx (4V)^2 = 144 MUJ` `:.` Loss in electrostatic energy `Delta u_i = u_i = (216 - 144) = 72 mu J` `rArr (Delta u) /u_1 = 72/216 = 1/3` |
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