- #1
moenste
- 711
- 12
Homework Statement
When a particular voltmeter of fixed resistance R, which is known to be accurately calibrated, is placed across the 1.8 kΩ resistor in the diagram below it reads 2.95 V. When placed across the 4.7 kΩ resistor it reads 7.70 V.
(i) Why do these two readings not add up to 12 V?
(ii) Calculate the resistance R of the voltmeter.
Answer: 10.3 kΩ.
2. The attempt at a solution
(ii) Let's find the current of the circuit: I = 12 / 4700 = 2.55 * 10-3 A. Then let's find the current of the 1800 Ω resistor: I1800 = 2.95 / 1800 = 1.63 * 10-3 A. So the resistance of the voltmeter is RV = VV / (I - I1800) = 2.95 / (2.55 * 10-3 - 1.63 * 10-3) = 3226.5 Ω. And it doesn't fit the answer...
I did the same calculation for the second position (voltmeter parallel to the 4700 Ω resistor) and got 1531.3 Ω.
What do I miss here? I consider the voltmeter to be connected in parallel. I also consider 2.95 V and 7.70 V as the reading of the voltmeter and the parallel circuit (so the 12 V is the same for both situations). Maybe the problem assumes that the EMF of 12 V changes to 2.95 V and 7.70 V respectively. Not sure.
P. S. This is a part of the problem, other parts are solved and I got the correct answers. But in case they are required for a better understanding, here is a image of the full problem.
Last edited: