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.

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

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

resistance.

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

To find the lamp's resistance, we can use Ohm's Law, which states that resistance (R) = voltage (V) / current (I). In this case, we have two batteries, each with an internal resistance, and the total resistance of the circuit is the sum of the internal resistance of both batteries and the lamp's resistance.

First, let's find the total voltage (V) across the lamp. Since the batteries are connected in series, their voltages add up. So, the total voltage (V) across the lamp is equal to the voltage of one battery (1.30 V) plus the voltage of the other battery (1.30 V), giving us a total voltage of 2.60 V.

The current (I) flowing through the lamp is given as 665 mA, which is equivalent to 0.665 A.

Next, we need to calculate the total internal resistance of the batteries. The internal resistances of the two batteries are given as 0.182 Ω and 0.139 Ω. Since the batteries are connected in series, their resistances also add up. Therefore, the total internal resistance (R_total) is equal to the sum of the two internal resistances, which is 0.182 Ω + 0.139 Ω = 0.321 Ω.

Now, we can determine the resistance of the lamp using Ohm's Law: R = V / I, where V is the total voltage (2.60 V) and I is the current (0.665 A). Therefore, the lamp's resistance (R_lamp) = 2.60 V / 0.665 A = 3.909 Ω.

Therefore, the lamp's resistance is approximately 3.909 ohms.