# How to derive equation for current from parallel batteries given internal resistance and load?

I'm studying DC circuits in physics right now and have come across this frustratingly unclear part of the textbook.

Apparently, the current through a circuit given ideal emf, internal resistance, and circuit load is $$\I = \frac{\epsilon}{r_{eq} + R}\$$ where $$\r_{eq} = (\frac{1}{r_1} + \frac{1}{r_2})^-1\$$

But the derivation of this equation is unclear, and to make matters worse. I'm pretty sure there's a typo because it says $$\IR = \frac{\epsilon}{r_{eq} + R}\$$ which doesn't make any sense.

So how does one get to the equation $$\I = \frac{\epsilon}{r_{eq} + R}\$$?

The derivation is correct under the assumption $$\\epsilon_1 = \epsilon_2 = \epsilon\$$, but in the voltage drop across the resistor there is indeed a missing $$\R\$$ on the right-hand side. The correct equation is

$$IR = \frac{R}{R+r_\mathrm{eq}}\epsilon\tag{1}$$

To find the current according to the book procedure, solve the equation at loop fcdef for $$\I_2\$$,

$$I_2 = \frac{\epsilon-IR}{r_2}$$

and substitute this into the equation at loop abcfa, obtaining

$$I_1 = \frac{r_2}{r_1}I_2 = \frac{\epsilon-IR}{r_1}$$

and finally substitute $$\I_1\$$ and $$\I_2\$$ into $$\I = I_1+I_2\$$ such that

$$I = (\epsilon-IR)\bigg(\frac{1}{r_1}+\frac{1}{r_2}\bigg)=\frac{\epsilon-IR}{r_\mathrm{eq}}.$$

Solving for $$\I\$$ yields the desired result

$$I = \frac{\epsilon}{R+r_\mathrm{eq}}.$$

For alternative solutions, the most straightforward way to derive that voltage is to use Millman’s theorem, which applied to that circuit yields

$$IR = \frac{\epsilon_1/r_1+\epsilon_1/r_2}{1/r_1+1/r_2+1/R}$$

Assuming $$\\epsilon_1 = \epsilon_2 = \epsilon\$$ and simplifying yields (1).

Another way is to use Thévenin’s theorem to find the Thévenin’s equivalent of the circuit part on the left of nodes d and e.