Phy Capacitor Charge Decay問題

2009-08-11 7:02 am

[img]http://img.photobucket.com/albums/v226/st2910/C12circuit.jpg[/img]
From the above circuit,
1) after the switch is closed for a long time, the capacitor is removed, is the ammeter reading affected? Why or why not?
2 after the switch is closed for a long time, the switch is opened. By recording the time when the pointer of the ammeter passes successive divisions of the scale and plotting a graph, I find that the half time of the decay is 36 seconds. However, theoratically, using the formula "Theoretical t1/2 = (ln 2) CR"(the equation is 100% correct), the half time calculated is 32.43 seconds. Why is there such an discrepancy?
Thank you for your help~

回答 (1)

Prof. Physics

2009-08-11 6:53 pm

✔ 最佳答案

1. The ammeter reading will not be affected. It is

I = V / R = 6 / 100 000 = 6 X 10-5 A

It is because when the switch is closed for a long time, the capacitor is fully charged, and it acts as an open circuit to the d.c. current. So, removing it does not affect the ammeter reading since it is already an open circuit.

2. Your equation is correct.

t1/2 = CR ln2

The reason for you to find the half time of the decay to be 36 seconds is because there exists stray capacitance in the surroundings, which enlarges the value of C and so the experimental half time is larger than that of theoretical.

Stray capacitance can be given by any metallic substance in the surroundings. For example, the water pipe in the laboratory, the lamp, the fan, e.t.c.

參考: Physics king

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