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 .

Explanation / Answer

I * r + I * R1 + I*R2 = V
When R2 = 22
V2 = 0.35 Vmax
V2 = 0.35 * 20
V2 = 7volt

V2 = I*R2
I = V2 / R2
I = 2 Amp

I * r + I * R1 + I*R2 = V
V - I*r = 7/22 * (20 + 22)
V - 7/22*r = 147/11 ------------1

20 + I1*r - V + I1*R1 = 0
20 + I1*r + I1*20 = V

When R2 will be very large, Current Flowing in the circut will be negligible and thus , Voltage drop in the Resistane R1 and Internal Resistance will be negligible.

Therefore,
V = 20 volt.
Substituing Value of V in eq 1

V - 7/22*r = 147/11
20 - 147/11 = 7/22 * r
r = 20.85
Internal resistance of the battery, r = 20.85

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