When no resistance is connected across a real battery, the potential difference across its terminals is measured to be 6V. When a R = 2Ω resistor is connected across the battery, a current of 2A is measured through the resistor. What is the internal resistance, r, of the battery?

1Ω

\(\frac{6}{r + 2 \Omega}\) = 2 A

r = 1 \(\Omega\)