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 = 15 ?, 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 = 10 V.

You then measure that ?V2 = 0.40 ?Vmax when R2 = 22 ?.

Explanation / Answer

current in the circuit = v2/r2 = 0.4 x 10/22 =0.18A

ohm law

10-ir-15i-22=0

r=10-4.86/.18

=28.555ohm

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