Concept explainers
Two 1.50-V batteries—with their positive terminals in the same direction—are inserted in series into a flashlight. One battery has an internal resistance of 0.255 Ω, and the other has an internal resistance of 0.153 Ω. When the switch is closed, the bulb carries a current of 600 mA. (a) What is the bulb’s resistance? (b) What fraction of the chemical energy transformed appears as internal energy in the batteries?
(a)
Answer to Problem 1P
Explanation of Solution
Given information: Internal resistance of one battery is
Explanation:
`
Write the expression for the total emf for the circuit.
Here,
Substitute
Thus, the total emf for the circuit is
Formula to calculate the total internal resistance for the circuit.
Here,
Substitute
Thus, the total internal resistance for the circuit is
Formula to calculate the value of the bulb resistance.
Here,
Substitute
Thus, the value of the bulb resistance is
Conclusion:
Therefore, the value of the bulb resistance is
(b)
Answer to Problem 1P
Explanation of Solution
Given information: Internal resistance of one battery is
Explanation:
Formula to calculate the internal energy of the battery.
Here,
Substitute
Thus, the internal energy of the battery is
Formula to calculate the fraction of the chemical energy transformed appears as internal energy in the batteries.
Here,
Substitute
Thus the fraction of the chemical energy transformed appears as internal energy in the batteries is
Conclusion:
Therefore, the fraction of the chemical energy transformed appears as internal energy in the batteries is
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Chapter 27 Solutions
Physics for Scientists and Engineers
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