P28_005.PDF

5. (a) Let i be the current in the circuit and take it to be positive if it is to the left in R 1 .Weuse

Kirchhoﬀ’s loop rule:

E 1 −

iR 2 −

iR 1 −E 2 =0.Wesolvefor i :

i =

R 1 + R 2 = 12V

6 . 0V

4 . 0Ω+8 . 0Ω =0 . 50 A .

−

A positive value is obtained, so the current is counterclockwise around the circuit.

(b) If i is the current in a resistor R , then the power dissipated by that resistor is given by P = i 2 R .

For R 1 , P 1 =(0 . 50 A) 2 (4 . 0Ω)=1 . 0 W and for R 2 , P 2 =(0 . 50 A) 2 (8 . 0Ω)=2 . 0W.

, then the battery supplies energy at the rate P = i

if the current and emf are in opposite directions. For

E 1 , P 1 =(0 . 50 A)(12 V) = 6 . 0W and

E 2 , P 2 =(0 . 50 A)(6 . 0V)=3 . 0 W. In battery 1the current is in the same direction as the emf.

Therefore, this battery supplies energy to the circuit; the battery is discharging. The current in

battery 2 is opposite the direction of the emf, so this battery absorbs energy from the circuit. It is

charging.

E 1 −E 2

provided the current and emf are in the same direction. The battery absorbs energy at the rate

P = i

for