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What is the difference between the ideal and practical ammeter?

To be more precise, here's my task:

task

I do not need a solution to this. Just looking for a clear explanation between practical and ideal ammeter, because I've only encountered the latter one so far.

Thanks.

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    \$\begingroup\$ What does "RA" represent? This symbol doesn't appear anywhere in the drawing. Is it a convention in your class that this is the internal resistance of the ammeter? (That's also a hint to you about the difference between real and ideal ammeters) \$\endgroup\$
    – The Photon
    Nov 29, 2018 at 17:32
  • \$\begingroup\$ @ThePhoton - I'm pretty sure that RA is the resistance of the ammeter. \$\endgroup\$ Nov 29, 2018 at 18:11
  • \$\begingroup\$ @WhatRoughBeast, that's my guess, but it should be stated explicitly. \$\endgroup\$
    – The Photon
    Nov 29, 2018 at 18:14

1 Answer 1

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Short answer is an ideal measurement instrument does not perturb the system it is measuring and produces a completely accurate rendition with infinite precision.

Real instruments always change the system to some degree and have finite sensitivity, accuracy, and precision and generally have some noise level.

In the case of a typical ammeter the primary issue is that it introduces additional resistance (RA in your example) as opposed to zero ohms into the circuit path. It will also have frequency limitations and will introduce additional parasitic components (capacitance, inductance) into the circuit that may change the performance of the circuit unintentionally. The latter generally is a problem in RF and sensitive circuit nodes.

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