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 = 10 V.

You then measure that ?V2 = 0.45 ?Vmax when R2 = 23 ?.

What is the internal resistance of the battery

Explanation / Answer

R1 = 20 ohm and R2 = 23 ohm
both the resistance are in series
When R2 is sufficiently large then the voltage drop is 10 V So we can say that the EMF of the battery will be 10 V.
Now when R2 = 23 ohm then voltage drop is 0.45*10 = 4.5 V
So current through the R2 , I = 4.5/23 = 0.195 A
since all the resistances are in series therefore current will be same in all the resistances.
So votlage drop across R1 = 0.195*20 =3.91 V
Total voltage drop across R1 and R2 = 3.91+4.5 = 8.41 volt
So voltage drop across battery = 10 -8.41 = 1.586 volt
So resistance V= ir
r = 1.586/0.195 =8.138 ohm

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