You have developed a new way to measure the internal resistance of a battery. Th

Question

You have developed a new way to measure the internal resistance of a battery. The circuit you have created is shown below: a non-ideal battery, with EMF (open-circuit voltage) and internal resistance r in the red box, a constant resistor R1-36.6 , and a variable resistor R2, R2 has a resistance you can change over a wide range of values, including very large resistances. n this circuit, you measure the voltage over R2.AV2. R1 is used to avoid damaging the battery due to high currents. In measuring a battery, you find the voltage V2 reaches a constant maximum value. AV,max = 19 V when R2 becomes very much larger than R1 and the internal resistance r You then reduce the resistance of R2 and when R2 = 40.5 , the voltage is reduced todVz" 0.48 Vmax. What is the internal resistance of the battery? Give your answer in Ohms to three significant digits. You won't be graded on the number of digits, this is just to avoid being counted wrong due to rounding. 6.061 V R2 EMF

Explanation / Answer

R2 is maximum so

dVmax = I1*R2 = E/(r+R1)

19 = E/(r+36.6) .................(1)

if R2 = 40.5 ohms

dV2 = I2*R2 = E/(r+R1+R2)

0.48*19 = E/(r + 36.6 + 40.5) .........(2)

solving 1 & 2

r = 0.785 ohms

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