Why can the internal resistance of the DMM can be determined without taking into account the 100 Ω resistor? Could this be done for a resistor of any resistance value? Explain your answer.

For resistances in parallel we know that the overall resistance will be smaller in the value than any individual resistance. so, in this case we are concerned about [tex]\frac{1}{R}[/tex] = [tex]\frac{1}{Rs}[/tex] ⁺ [tex]\frac{1}{Rp}[/tex]

where Rs is the series resistor and Rp is the parallel resistor.

so, R = 0.9M, roughly.

So, as the infernal resistance is very high as compared to the resistance conntected in parallel that is 100 Ω is not be used.

okpalawalter8 okpalawalter8 · Answer 2 · 2022-03-21T16:05:04+01:00

The digital multimeter (DMM) can have its internal resistance determined without taking into account the 100 Ω resistor because the internal resistance is very high when compared to that of the parallel connection, and this does not change as the resistance principle remains the same.

What is the reason why the internal resistance of the DMM can be determined without taking into account the 100 Ω resistor?

Generally, Singles connection of resistors when connected, it topples the resultant parallel connection of resistors.

Therefore

[tex]\frac{1}{R} = \frac{1}{Rs} * \frac{1}{Rp}[/tex]

Hence, because the internal resistance (R)is very high when compared to that of the parallel (Rp), the 100ohms is not put to use.

In conclusion, This could be done for a resistor of any resistance value as the resistance principle remains the same.