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 = 20 V.

You then measure that V2 = 0.35 Vmax when R2 = 22 .

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

Explanation / Answer

Req of circuit R = R1 + R2 + r

when R2 is very large R is approamitely equal to R2.

i = e /R2

then voltage drop across R2 = IR = (e /R2) *R2 = 20 V

e = 20 V

When R2 = 22 ohm,

Req =(r + R1 + R2) = r + 22 +20 = 42 + r

V2 = iR2 = (20 / 42+r) 22 = 0.35 x 20

42 +r = 22/0.35

r =20.86 ohm

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