You find an old battery with an unknown EMF, \varepsilonε, and an unknown internal resistance, r. You can characterize the battery using two known resistors and a good ammeter. When you put the battery in series with the first resistor R_1 = 100~\OmegaR 1 =100 Ω, you measure a current through it of I_1 = 0.093 AI 1 =0.093A. You then remove the first resistor and replace it with the second resistor, R_2 = 500~\OmegaR 2 =500 Ω, and now measure a current through it of I_2 = 0.019I 2 =0.019. Calculate the internal resistance of this mystery battery. Assume the ammeter is ideal and has negligible resistance)

   r = 2.70 Ω

Explanation:

The voltage is given by the ratio

          V = i R

In the given circuit the external resistance is placed in series, so the voltage ratio is

         V = i₁ (R₁ + r)

         V = i₂ (R₂ + r)

Since the battery is the same, the voltage is the same in both cases

         i₁ (R₁ + r) = i₂ (R₂ + r)

         r (i₁ –i₂) = i₂ R₂ - i₁ R₁

         r = (i₂ R₂ –i₁ R₁) / (i₁ -i₂)

Let's calculate

          r = (0.019 500 - 0.093 100) / (0.093 - 0.019)

          r = 0.2 /0.074

         r = 2.70 Ω

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