Someone did an experiment with an ideal voltmeter and an ideal ampermeter using the circuit in the picture, and got the graph in the picture.
-We know use an unideal voltmeter, how will the graph change?
-We know use an unideal ampermeter (the voltmeter is ideal), how will the graph change?
The Attempt at a Solution
The graph basically displays u=ε-Ir.
I figured that if the voltmeter is not ideal, the same current measured by the ampermeter will be "shared" by the voltmeter and the resistor, causing a higher value for the same current. In the answers they say it won't change.
Then, I figured that if the ampermeter is not ideal, the value will decrease quicker, becuase now u=ε-Ir-IR, this one doesn't have an answer.
Where is my mistake? Am I right?
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