Two 1.30-V batteries-with their positive terminals in the same direction-are inserted in series into the barrel of a flashlight. One battery has an internal resistance of 0.182Ω, the other an internal resistance of 0.139Ω. When the switch is closed, a current of 665mA occurs in the lamp. What is the lamp's resistance? Answer in units of ohms.

Answer 1

R=E/I = 1.30/.665 = 1.955Ω

Of that, 0.182+0.139=0.321Ω is in the batteries.

So, the lamp contributes only 1.634Ω

Answer 2

R = E/I = 2.60/0.665 = 3.91 Ohms = Total
resistance.

Rl = 3.91 - 0.182-0.139 = 3.589 Ohms =
Lamp resistance.