In Example, how will the percentage of error change if the ammeter’s internal resistance i...

In Example, how will the percentage of error change if the ammeter’s internal resistance increases to Rg = 15 Ω?

EXAMPLE

Instrument Reading Error

In Figure, determine the error percentage of a current reading due to the insertion of an ammeter. R3= R2= 100 Ω, R1 = 200 Ω, and the internal resistance of the ammeter is Rin = 5 Ω.

FIGURE Circuit for Example

SOLUTION

Before inserting the ammeter, the total current flowing through the 10 V source is calculated to be:

Because the two parallel resistors, R3 and R2, are equivalent, the current passing through R3 is 0.04/2 = 0.02 A.

After inserting the ammeter, due to the resistance of the ammeter, the final resistance of the two parallel branches will not be equivalent. In this case, the total current flowing through the 10 V resistor will be:

In this case, using the current division concept, the current flowing through R3 is:

Therefore, the error percentage is:

Thus, the error percentage due to the current reading corresponds to 3%.