P28 071

background image

71.

(a) By symmetry, we see that i

1

is half the current that goes through the battery. The battery current

is found by dividing

E by the equivalent resistance of the circuit, which is easily found to be 6.0 Ω .

Thus,

i

1

=

1

2

i

bat

=

1

2

12 V

6.0 Ω

= 1.0 A

and is clearly downward (in the figure).

(b) We use Eq. 28-14: P = i

bat

E = 24 W.


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