Recent content by killdevil

A voltage divider? I'm not sure I follow what you mean by that. Many thanks for your help so far.

After substituting EMF - Ir based on the definition of internal resistance, I arrive at a ratio that looks like this: P(1)/P(2) = ((V[o]^2)(R) + (V[o]^2)(r)) / ((r)(V[o]-Ir)^2) Which expands out into nastiness if I do the exponential expansion... the problem is that I've got current in the...

Also, I am not at all certain how to extrapolate from this in order to set up a ratio between r and R. Oh how I wish my professor spoke English...

Okay... the power is V^2/R because the circuit is a series circuit? And likewise, the total power dissipated in both resistors has to be P=V^2/(R+r) because resistances just add in a series situation... To generalize... would it be reasonable to say that you always want to try to analyze...

1. The problem: A resistor has a resistance R, and a battery has an internal resistance r. When the resistor is connected across the battery, ten percent less power is dissipated in R than would be dissipated if the battery had no internal resistance. Find the ratio r/R. 2. Homework...