Question

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)

Answer 1

r = 2.70 Ω

Step-by-step 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