標題:
Physics---Capacitance 2
發問:
Two capacitors of 10μF and 47μF are initially charged to 9V and 6V respectively . They are then connected together .Find the p.d. across each capacitor and the charge on each capacitor if a. they are connected with like plates together ,b.they are connected with unlike plates together .In each case... 顯示更多 Two capacitors of 10μF and 47μF are initially charged to 9V and 6V respectively . They are then connected together .Find the p.d. across each capacitor and the charge on each capacitor if a. they are connected with like plates together , b.they are connected with unlike plates together . In each case calculate the energy dissipated in the circuit .
最佳解答:
(a) With the like plates together, the charge amount is conserved since there's no neutralization occuring. Therefore, Charge of the 10μF cap initially = 10 10-6 9 = 9 10-5 C Charge of the 47μF cap initially = 47 10-6 6 = 2.82 10-4 C So total charge = 3.72 10-4 C Now, since they are connected together, they will have the same p.d., say V and hence: (47 10-6 + 10 10-6) V = 3.72 10-4 V = 6.53 V And the charges are: 10μF cap = 10 10-6 6.53 = 6.53 10-5 C 47μF cap = 47 10-6 6.53 = 3.07 10-4 C (b) If the unlike plates are connected, neutralzation occurs and then net charge remained = 2.82 10-4 - 9 10-5 = 1.92 10-4 C So using the same method as (a): (47 10-6 + 10 10-6) V = 1.92 10-4 V = 3.37 V And the charges are: 10μF cap = 10 10-6 3.37 = 3.37 10-5 C 47μF cap = 47 10-6 3.37 = 1.58 10-4 C 2008-02-28 14:57:03 補充： Initial energy: 0.5 × (10^-5 × 9^2 + 4.7 × 10^-5 × 6^2) = 1.251 × 10^-3 J (a) Final energy = 0.5 × (10^-5 × + 4.7 × 10^-5) × 6.53^2 = 1.215 × 10^-3 J Energy dissipation = 3.573 × 10^-5 J (b) Final energy = 0.5 × (10^-5 × + 4.7 × 10^-5) × 3.37^2 = 3.237 × 10^-4 J Energy dissipation = 9.273 × 10^-4 J
其他解答:
Physics---Capacitance 2
發問:
Two capacitors of 10μF and 47μF are initially charged to 9V and 6V respectively . They are then connected together .Find the p.d. across each capacitor and the charge on each capacitor if a. they are connected with like plates together ,b.they are connected with unlike plates together .In each case... 顯示更多 Two capacitors of 10μF and 47μF are initially charged to 9V and 6V respectively . They are then connected together .Find the p.d. across each capacitor and the charge on each capacitor if a. they are connected with like plates together , b.they are connected with unlike plates together . In each case calculate the energy dissipated in the circuit .
最佳解答:
(a) With the like plates together, the charge amount is conserved since there's no neutralization occuring. Therefore, Charge of the 10μF cap initially = 10 10-6 9 = 9 10-5 C Charge of the 47μF cap initially = 47 10-6 6 = 2.82 10-4 C So total charge = 3.72 10-4 C Now, since they are connected together, they will have the same p.d., say V and hence: (47 10-6 + 10 10-6) V = 3.72 10-4 V = 6.53 V And the charges are: 10μF cap = 10 10-6 6.53 = 6.53 10-5 C 47μF cap = 47 10-6 6.53 = 3.07 10-4 C (b) If the unlike plates are connected, neutralzation occurs and then net charge remained = 2.82 10-4 - 9 10-5 = 1.92 10-4 C So using the same method as (a): (47 10-6 + 10 10-6) V = 1.92 10-4 V = 3.37 V And the charges are: 10μF cap = 10 10-6 3.37 = 3.37 10-5 C 47μF cap = 47 10-6 3.37 = 1.58 10-4 C 2008-02-28 14:57:03 補充： Initial energy: 0.5 × (10^-5 × 9^2 + 4.7 × 10^-5 × 6^2) = 1.251 × 10^-3 J (a) Final energy = 0.5 × (10^-5 × + 4.7 × 10^-5) × 6.53^2 = 1.215 × 10^-3 J Energy dissipation = 3.573 × 10^-5 J (b) Final energy = 0.5 × (10^-5 × + 4.7 × 10^-5) × 3.37^2 = 3.237 × 10^-4 J Energy dissipation = 9.273 × 10^-4 J
其他解答:
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