Problem 21.26

21.26
The total resistance is given by
$R = \frac{V}{I}$
$R = \frac{3V}{.6 A}$
The resistance of the lamp, since it is a series circuit is given by
R_lamp = R - r_batteries
$R_{lamp} = 5 \Omega - .41 \Omega ~=~ 4.59 \Omega$

b)
The fraction of power dissipated in the betteries is given by
$f = \frac{P_{batteries}}{P_{total}}$
$f = \frac{(.408 \Omega)I^{2}}{(5 \Omega)I^{2}}$
f = .0816, or 8.16%