Forum Discussion
jrnymn7
Oct 06, 2014Explorer
V_ps = V_bat + I_cc * R_cable
When in cc, the supply voltage will always be a constant (I_cc * R_cable) above the V_bat.
so, batts are at 12.2v...
12.2v + (20 x .015) = 12.2v + 0.3v = 12.5v
The psu's output voltage would start at 12.5v and rise accordingly... once batts were at say 12.6v the psu would be outputting 12.9v ?
When in cc, the supply voltage will always be a constant (I_cc * R_cable) above the V_bat.
so, batts are at 12.2v...
12.2v + (20 x .015) = 12.2v + 0.3v = 12.5v
The psu's output voltage would start at 12.5v and rise accordingly... once batts were at say 12.6v the psu would be outputting 12.9v ?
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