electrons to miliamps converter

[RITESH KAKKAR]

Hello,

If 0.883x1017 electrons flow through a conductor in 239 ms, what is the current in amperes?
59.11mA
59.1kA
59.11A
59.11microA

I have done this 1Amp 1 Coulomb 1sec = 6x10^18

1000mA 1000mSec = 6x10^18
1mA 1mSec = 6x10^18 / 1000000

so, for 239mSec =6x10^18 / 1000000 * 239

how to calculate?

 

[FvM]

Referring to elementary charge constant e

I = 0.883x10^17 / 0.239 * 1.602*10^-19

 

[RITESH KAKKAR]

can you explain this?

- - - Updated - - -

What is the voltage between two points if 55 mJ of energy are required to move 40x1017 electrons between the two points?
75.8mV
85.8mV
95.8mV
65.8mV

 

[FvM]

E = Q * U

put in and solve respectively. Use elementary charge constant as previously.
You mean to write 40*10^17 not 40*1017.

 

[RITESH KAKKAR]

Referring to elementary charge constant e

I = 0.883x10^17 / 0.239 * 1.602*10^-19

I dnt understand this.

 

[FvM]

0.883x10^17 / 0.239 is count of electrons per second

multiply with elementary charge constant (= the charge of 1 electron) to get Q/s (= A).

 

[RITESH KAKKAR]

but is in 239 ms
1C and 1 A is in Sec

i dnt get you sir

 

[FvM]

Consider 239 ms = 0.239 s

See also https://en.wikipedia.org/wiki/Metric_prefix

 

[RITESH KAKKAR]

but 1sec contain 6.25x10^18 for 1Amp 1 C

 

[FvM]

but 1sec contain 6.25x10^18 for 1Amp 1 C

Why but? That's just the inverse of elementary charge.

1/e = 6.25x10^18

I would prefer to use the natural constants that can be looked up in a text book, but if you like, write

I = 0.883x10^17 / 0.239 / 6.25x10^18

Same result, 0.059 A = 59 mA

 

[RITESH KAKKAR]

I = 0.883x10^17 / 0.239 / 6.25x10^18

Same result, 0.059 A = 59 mA

the unit of amp mean in sec but the current is flowing in millisecond will be same?

 

[FvM]

Review the previous calculation. I did it in seconds and amps.