Find the resistance of the lamp

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The discussion revolves around calculating the resistance of a lamp powered by two 1.50 V batteries in series, each with specified internal resistances. To find the lamp's resistance, Ohm's Law (V = IR) is applied, where the total voltage and current are known, along with the sum of the internal resistances. The second part of the problem involves determining the fraction of chemical energy converted to internal energy within the batteries, requiring the calculation of energy consumed by the internal resistance. Participants express uncertainty about the initial steps and the interpretation of the second question. The conversation emphasizes the need for clarity in applying electrical principles to solve the problem effectively.
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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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