38.
(a)
E = V + ir = 12 V + (10 A)(0.050 Ω) = 12.5 V.
(b) Now
E = V
+ (i
motor
+ 8.0 A)r, where V
= i
A
R
light
= (8.0 A)(12 V/10 A) = 9.6 V. Therefore,
i
=
E − V
r
− 8.0 A =
12.5 V
− 9.6 V
0.050 Ω
− 8.0 A = 50 A .