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 = 20 ?, 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= 15 V.

You then measure that ?V2 = 0.40 ?Vmax when R2 = 25 ?.

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

Explanation / Answer

when R2 become larger then current flow through circuit will be zero so voltmeter actually measuring the actual voltage of battery

Vbattery    = 15 V

When R2 = 25 then

  0.40 Vmax   = I*R2 = ( Vbattery / Rtotal ) *R2   = (15 V /R1+r+R2) * R2 =  0.40 Vmax

( 15/ 20+25+r ) *25 = 0.40*15

from here we found

r = 17.5

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