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a.) The resistance of the bulb is calculated l to be 4.58 Ohm and b.) The fraction of the chemical energy transformed which appears as internal energy in batteries is 8.4%.
Energy transfer to an electric circuit per unit of electric charge measured in volts is called electromotive force .
Given, Voltage= 1.5 Volts
Internal resistance A is 0.240 Ohms
Internal resistance B is 0.180 Ohms
and current = 600 mA = 0.6 Amps
a. To find resistance of the bulb:
The total electromotive force (E) of the electric circuit:
E = 2*1.5
= 3.0 Volts
Total internal resistance = 0.240 + 0.180 = 0.42 ohms
Electromotive force (E) is given by the formula:
E= V+ I r
E is electromotive force
V is voltage and I is the current.
r is internal resistance.
Ohm's law is given by:
V= I R
Given, Resistance, R = 4.58 Ohms
E = IR + I r
R= E/I - r
= (3/0.6) - 0.42
= 4.58 ohms
b. To determine what fraction of chemical energy transformed appears as internal energy in batteries:
the electromotive force (E) in the batteries. E b = I R
= 0.6 * 0.42
= 0.252 Volt
Percent= (Eb / E) * 10
=0.252/3
percent = 8.4 %
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Question : 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? %