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| 1. |
Calculate the potential difference and the energy stored in the capacitor C_2 in the circuit shown in the Fig. Given potential at A is 90 V , C_1 = 20 mu F , C_2 = 30 mu Fand C_3 = 15 mu F |
| Answer» <html><body><p></p>Solution :Let as <a href="https://interviewquestions.tuteehub.com/tag/shown-1206565" style="font-weight:bold;" target="_blank" title="Click to know more about SHOWN">SHOWN</a> in Fig `V_1 , V_2` and `V_3` be the respective potential differences across the capacitors `C_1 , C_2` and `C_3` respectively . In series combination same charge Q exists on each capacitor and `V_1 : V_2 : V_3 = (1)/(C_(1)) : (1)/(C_(2)) : (1)/(C_(3))` <br/> `implies V_(1)= (C_(2))/(C_(1)) V_(2) = (30)/(20) V_(2)` and `V_(3) = (C_(2))/(C_(3)) V_(2) = (30)/(15) V_(2)` <br/> But `V_(1) + V_(2) + V_(3) = 90 V` , hence we have <br/> `(30)/(20) V_(2) + V_(2) + (30)/(15) V_(2) = 90 V implies 45 V_(2) = 90 V implies V_(2) = 20 V ` <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/U_LIK_SP_PHY_XII_C02_E10_046_S01.png" width="80%"/> <br/> Moreover , energy stored in capacitor `C_2` is `V_2 = (1)/(2) C_(2) V_(2)^(2) = (1)/(2) xx (30 mu F) xx (20 V)^(2) = <a href="https://interviewquestions.tuteehub.com/tag/6000-329798" style="font-weight:bold;" target="_blank" title="Click to know more about 6000">6000</a> mu <a href="https://interviewquestions.tuteehub.com/tag/j-520843" style="font-weight:bold;" target="_blank" title="Click to know more about J">J</a> = 6.0 mJ`</body></html> | |