Daniel B. answered 02/28/22
A retired computer professional to teach math, physics
1.
Suppose that you have a circuit with resistance R,
and before inserting the ammeter, there is current I.
That implies voltage IR.
After inserting an ammeter with resistance r,
the total resistance of the circuit becomes R+r, and
the current becomes IR/(R+r).
That causes the ammeter to measure current different from the original I,
by a factor R/(R+r).
The smaller the r, the closer is R/(R+r) to 1, getting a closer approximation to I.
2.
Suppose you want to measure voltage across a resistance R,
and suppose the current flowing through it is I.
That means, the voltage across the resistance is IR.
After adding the voltmeter with resistance r in parallel to R,
the new resistance becomes
Rr/(R+r),
and the voltmeter will measure voltage IRr/(R+r).
That causes the voltmeter to measure voltage different from the original IR
by a factor r/(R+r).
The bigger the r, the closer is r/(R+r) to 1, getting a closer approximation to IR.
Mike R.
Thank you!02/28/22