You are working on developing a way of testing the internal resistance of a batt

Question

You are working on developing a way of testing the internal resistance of a battery. The circuit you have created is shown above, with a battery (EMF and internal resistance r in the red box), a constant external resistor R1 = 19 ?, and a variable resistor R2. R2 has a resistance you can change over a wide range of values. You measure the voltage over R2,?V2. The first resistor, R1, is there so you don't damage the battery by short-circuiting it.

In your measurements, you find that when R2 becomes very much larger than R1, the voltage ?V2 approaches a constant value, ?Vmax = 25 V.

You then measure that ?V2 = 0.50 ?Vmax when R2 = 21 ?.

What is the internal resistance of the battery? Give your answer in Ohms to three significant digits.

Explanation / Answer

e-(i*r) = 25 V

e-ir-iR1-iR2 = 0

i*R2 = 0.5*25= 12.5 V

i = 12.5/21 = 0.595 A

e-(0.595*r) = 25

12.5-(0.595*r) = 25

0.595*r = 12.5

r = 12.5/0.595 = 21 ohm

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