P28 038

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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

motor

=

E − V



r

8.0 A =

12.5 V

9.6 V

0.050 Ω

8.0 A = 50 A .


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