1. A rechargeable nattery rated at 1800 mA h means that it can produce a current of 1800 mA for 1 hr , or 1 mA for 1800 hr, and so on. How many coulombs of charge can be driven to flow through the battery?
2. How many electrons pass through a cow if it touches an electrified fence and receives a current pulse of 20 mA( on average) for 0.1 s?
物理 E and M
2011-09-29 6:53 pm
回答 (2)
2011-09-29 7:48 pm
✔ 最佳答案
1. No. of charge, Q = current X timeChange the units to S.I. units first.
Current = 1 mA = 0.001 A
Time = 1800 hr = 1800 X 3600 = 6 480 000 s
So, no. of charge = 0.001 X 6 480 000 = 6480 C
2. Again, no. of charge = current X time
Current = 20 mA = 0.02 A
time, t = 0.1 s
So, no. of charge = 0.02 X 0.1 = 0.002 C
But in 1 C, there are 6.25 X 10^18 electrons
So, no. of electrons = 0.002 X (6.25 X 10^18) = 1.25 X 10^16
參考: Prof. Physics
2011-09-29 7:52 pm
1. Since 1 A = 1 coul/s
hence,1 mA = 0.001 coul/s
1800 mA = 1800 x 0.001 coul/s = 1.8 coul/s
Therefore, 1800 mA-h = 1.8 x (3600) coul = 6480 coul.
2. Amount of charge passing through the cow
= (20 x 10^-3) x 0.1 coul = 2 x 10^-3 coul
Since an eectron carries 1.6x10^-19 coul of charge,
no. of electrons involved = 2x10^0-3/1.6x10^-19 = 1.25 x 10^16
hence,1 mA = 0.001 coul/s
1800 mA = 1800 x 0.001 coul/s = 1.8 coul/s
Therefore, 1800 mA-h = 1.8 x (3600) coul = 6480 coul.
2. Amount of charge passing through the cow
= (20 x 10^-3) x 0.1 coul = 2 x 10^-3 coul
Since an eectron carries 1.6x10^-19 coul of charge,
no. of electrons involved = 2x10^0-3/1.6x10^-19 = 1.25 x 10^16
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