InterviewSolution
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InterviewSolution
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Three identical capacitors `C_(1) , C_(2)` and `C_(3)` of capacitance `6 mu F` each are connected to a `12 V` battery as shown in Fig. Find (i) charge on each capcitor (ii) equivalent capacitance of the network. (iii) energy stored in the network of capacitors. |
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Answer» (i) Here, `V = 12V` and `C_(1) = C_(2) = C_(3) = 6mu F = 6 xx 10^(-6)F` Charge on capacitor `C_(3)` is `q_(3) = C_(3)V = 6 xx 10^(-6) xx 12 = 72 xx 10^(-6) = 72 mu C` Since, capacitor `C_(1) and C_(3)` are in series `:.` Equivalent capacitance `(1)/(C_(S)) = (1)/(C_(1)) + (1)/(C_(2)) = (1)/(C_(S)) = (1)/(6) + (1)/(6) = (2)/(6) = (1)/(3)` `:. C_(S) = 3 mu F` Charge on capacitor `C_(1) and C_(2)` is `q = C_(S) V = 3 xx 10^(-6) xx 12` `= 36 xx 10^(-6) = 36 mu C` `:.` Charge on each capacitor `C_(1) and C_(2) " is " 36 mu C` (ii) Since, `C_(1) and C_(2)` is in series `:.` Equivalent capacitance `C_(S) = 3mu F` Now `C_(3) and C_(S)` are in parallel `:.` Equivalent capacitance `C = C_(3) + C_(S)` `= 6 + 3 = 9mu F` (iii) Energy stored `= (1)/(2) CV^(2) = (1)/(2) xx 9 xx 10^(-6) xx (12)^(2) = (1)/(2) xx 9 xx 10^(-6) xx 144` `= 648 xx 10^(-6) = 6.48 xx 10^(-4) J` |
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