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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}\$?

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TheRealNoofle
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1 Answers1

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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.

Massimo Ortolano
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