AL phy problem capacitor

2009-06-12 9:02 pm
capacitor charging入面
Q=Qo(1-e^-t/RC)
RC越細,charging time就越快
咁即係resistance細就可以charge快d
叉得咁快對capacitor有咩副作用

另外一條
點解有時係d式,無論Q好,I好,V好
乘1-e^-t/RC
有時就冇左1-
淨係乘 e^-t/RC??
咩情況用唔同既式

回答 (1)

2009-06-13 4:04 am
✔ 最佳答案
叉得咁快對capacitor有咩副作用

In theory, there is no ill effect on the capacitor.
點解有時係d式,無論Q好,I好,V好
乘1-e^-t/RC
有時就冇左1-
淨係乘 e^-t/RC??
咩情況用唔同既式
The factor (1-e^-t/RC) denotes an exponential increase, whereas the facotr ( e^-t/RC) denotes an exponetial decrease.
For the case of a capcitor during charging, the charge on and hence voltage of the capacitor follows an exponential increasing manner. But the charging current decreases exponentially.
During discharging, all quantities, i.e. charge, voltage and current, decrease in an exponential manner.



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