2 Four $1.5\U{V}$ cells are connected in series to a $12\U{\Ohm}$
3 lightbulb. If the resulting current is $0.45\U{A}$, what is the
4 internal resistance of each cell, assuming they are identical and
9 This is simply an application of the procedure outlined in Question 13.
10 The external resistance is the lightbulb $R_{ext}=12\U{\Ohm}$.
11 The total internal resistance is the sum of all the individual cell resistances $r_{int}=6r$.
12 The total voltage is the sum of all the individual cell voltages $V = 6\cdot 1.5\U{V} = 9\U{V}$.
13 Putting these together we have
15 r_{int} = 6r &= \frac{V}{I} - R_{ext} = \frac{9\U{V}}{0.45\U{A}} - 12\U{\Ohm} \\
16 r &= \ans{1.3\U{\Ohm}}