# Algorithm for “short-circuit” current draw

Alrighty let's pretend that the circuit contains no resistance other than the battery internal resistance and the wire's resistance itself,

Would this be a correct algorithm? (Using OHM'S law)

BR - Battery internal resistance
WR - Wire Resistance


V / (BR + WR) = A

I am just trying to understand the basics before diving into more of the mechanics of our crazy world.

• You've got the division upside-down. Voltage should be on top. – Dan Laks Jan 26 '17 at 4:28
• Yep, that's ohms law! You got it. You won't have a lot of resistance though, so that means a LOT of current. Not so good an idea, but a great place to start. – Brian Dohler Jan 26 '17 at 4:33
• It's "I" not "A". "I" is the formulaic representation of current and a current of 5 amps would also be called 5 A. You did say you were trying to understand the basics so, best learn the proper way to write things before they become a bad habit. I would also call the resistors RB and RW (or Rb and Rw or better still $R_B$ and $R_W$. – Andy aka Jan 26 '17 at 8:18