P28_009.PDF

9. (a) If i is the current and ∆ V is the potential diﬀerence, then the power absorbed is given by P = i ∆ V .

Thus,

∆ V = P

= 50W

1 . 0A =50V .

Since the energy of the charge decreases, point A is at a higher potential than point B; that is,

V A −

V B =50V.

(b) The end-to-end potential diﬀerence is given by V A −

V B =+ iR +

,where

is the emf of element

C and is taken to be positive if it is to the left in the diagram. Thus,

= V A −

V B −

iR =

(1 . 0A)(2 . 0Ω)=48V.

−

, so it is toward the left. The negative terminal is at B .

50V