Resistance (R) = $\frac{V}{I} = \frac{200}{20} = 10 \, \Omega$
$\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I}$
$\frac{\Delta R}{10} = \frac{4}{200} + \frac{0.2}{20} = 0.02 + 0.01 = 0.03$
$\Delta R = 0.3 \, \Omega$
Thus, R = (10 ± 0.3) Ω


A battery of emf \( E \) and internal resistance \( r \) is connected to a rheostat. When a current of 2A is drawn from the battery, the potential difference across the rheostat is 5V. The potential difference becomes 4V when a current of 4A is drawn from the battery. Calculate the value of \( E \) and \( r \).