Solve Internal Resistance Homework Problem

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SUMMARY

The problem involves a 6.0V battery connected to a parallel combination of a 10 ohm and a 40 ohm resistor, resulting in a total current of 0.25A. The internal resistance (r) is calculated using the equation E = IR + Ir. The combined resistance of the parallel resistors is 8 ohms, leading to the conclusion that the internal resistance must be 16 ohms, which contradicts the answer book stating it should be 4 ohms. The correct approach confirms that R + r equals 24 ohms, indicating a miscalculation in the initial assumptions.

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  • Understanding of Ohm's Law (E = IR)
  • Knowledge of parallel resistor combinations
  • Familiarity with internal resistance concepts
  • Ability to perform basic algebraic manipulations
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Homework Statement



"A 6.0V battery is connected across a parallel combination of two resistors ; one 10 ohm and one 40 ohm. Thee total current provided by the battery is 0.25A. Obtain a value for the internal resistance.


Homework Equations



E= IR +Ir

The Attempt at a Solution



I find the combined resistance of the resistors to be 8 ohms, and sub this into the equation above, and solve for little r, but get an answer of 16 ohms; the correct answer, according to the answer book is 4 ohms. Please can you show me a correct solution to the question!

Thanks
 
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Your answer is correct. Since 6.0 V produces 0.25 A of current, R+r must be 24 ohms.
10 and 40 ohms parallel is indeed 8 ohms, so r must be 16 ohms.
 

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