# Electrical Current: hour rate

1. Jun 23, 2009

### Effort

I am trying to set up intial conditions for an FEA simulation and I need to know the current density. I first need to get the current which is specified as a"constant current of 0.51 h rate." This is a disharge rate. I assumed that the h meant hour. Can I derive the currnet form that information?

Put another way, the paper said that the discharge occurs in steps of 1.1 (microAh) / ((cm^2)*(micrometer)) for 25 seconds and then maintianting an open circuit for 75 seconds. I am assuming this is just saying the 0.51 h rate another way. Can someone explain how I relate the two?

I used the 1.1 (microAh) / [(cm^2)*(micrometer)] to calculate the currnet density in A/m^2 by

[1.1 microAhr / ((10^-2)^2*(10^-6)) ] * [depth=10*10^-3] * [3600sec/hr * 1/25sec.]

= 15840 A/m^2.....this seemed kind of high, but I'm not an EE.

Any insights would be appreciated

2. Jun 23, 2009

### negitron

Looks about right to me at a cursory glance. 15,840 A/m^2 would only put about 80 mA through, say, 10 AWG wire which is rated by the NEC for use on 30 A circuits. So, not all that high, really.