Find the resistance of the lamp

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SUMMARY

The discussion focuses on calculating the resistance of a lamp connected to two 1.50 V batteries in series, with internal resistances of 0.255 ohms and 0.153 ohms. Using Ohm's Law (V = IR), the total voltage (3.0 V) and the total internal resistance (0.408 ohms) allow for the determination of the lamp's resistance when a current of 600 mA is flowing. Additionally, the discussion addresses the fraction of chemical energy converted to internal energy in the batteries, emphasizing the need to compute energy consumed by the internal resistance.

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  • Understanding of Ohm's Law (V = IR)
  • Basic knowledge of series circuits and internal resistance
  • Familiarity with electrical energy concepts
  • Ability to perform calculations involving voltage, current, and resistance
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  • Calculate the lamp's resistance using Ohm's Law with given values
  • Learn about energy conservation in electrical circuits
  • Explore the concept of internal resistance in batteries
  • Investigate the relationship between current, voltage, and power in circuits
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Homework Statement


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 0.255 ohms, the other has a resistance of 0.153 ohms. When the switch is closed, a current of 600 mA occurs in the lamp. (a) What's the lamps resistance? (b) What fraction of the chemical energy transformed appears as internal energy in the batteries?


Homework Equations





The Attempt at a Solution


I'm not sure how to start the problem. Please Help.
 
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a) By Ohm's Law, V = IR. So you know I, you know V, R is the sum of the two internal resistances and the flashlight so it can be easily solved.

b) Not sure what this is asking exactly.
 
For the last part compute the energy consumed by the interal resistance.
 

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