here is the circuit.
I said that r (internal resistnce of battery) decreases because of the resistors in parallel rule. I also said that the voltmeter with low resistance draws more current.
are these correct?
am i right in thinking i can use the resistors in parallel equation rule to say that the value for the internal resistance has decreased????Draw the circuit with real models for the meters and then figure out what the E and r values would be if the meter readings were taken at face value. How do those compare to the actual values? Which meter's non-ideal resistance dominates the error?
No, because the resistors aren't in parallel. In order to be in parallel, they have to have the same voltage across each. Is that the case here?am i right in thinking i can use the resistors in parallel equation rule to say that the value for the internal resistance has decreased????
I meant the voltmeter and little r(internal resistance)? does the 1/v=1/v(1)+1/v(2) apply. as in the exam i said the resistance of r falls if the voltmeter resistance is low.No, because the resistors aren't in parallel. In order to be in parallel, they have to have the same voltage across each. Is that the case here?
I think they aren't in parallel. Because in addition r there is also ε in series with it.I meant the voltmeter and little r(internal resistance)?
What do you mean by this? Do you mean 1/R = 1/R1 + 1/R2?does the 1/v=1/v(1)+1/v(2) apply.