posted by Selma .
A voltmeter of internal resistance of 5000 Ω is connected across the poles of a battery of internal resistance 0.20 Ω. The voltmeter reads 1.4993 volts. What is the actual zero-current emf of the battery?
Would this formula be I = V/(r+R) = 1.4993/(5000+0.20)?
I (r+R) = V, where V is the zero-current emf. I is the current when the measurement is made. R is the voltmeter's internal (series) resistance. r is the battery's internal resistance.
The measured voltage is I R = V' = 1.4933
V = (V'/R)*(r+R) = 1.4933*(5000.2/5000)
which is not much different.