Two 1.50-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 0.240 ω, the other an internal resistance of 0.180 ω. when the switch is closed, a current of 600 ma occurs in the lamp. (a) what is the bulb's resistance? ω (b) what fraction of the chemical energy transformed appears as internal energy in the batteries? %

a. The resistance of the bulb is equal to 4.58 Ohms.

b. The fraction of the chemical energy transformed which appears as internal energy in the batteries is 8.4%.

Given the following data:

a. To find the resistance of the bulb:

First of all, we would determine the total electromotive force (E) of the electric circuit:

E = 3.0 Volts

Total internal resistance =

Mathematically, electromotive force (E) is given by the formula:

   ....equation 1.

Where:

According to Ohm's law, voltage is given by:

   ....equation 2

Substituting eqn 2 into eqn 1, we have:

Resistance, R = 4.58 Ohms

b. To determine what fraction of the chemical energy transformed appears as internal energy in the batteries:

First of all, we would determine the electromotive force (E) in the batteries.

= 0.252 Volt

Percent = 8.4 %

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