I'm required to take the limit of the voltage division equation(internal voltmeter resistance included as Rm) when Rm goes to infinity to show that the limit equals to the basic voltage division equation.

\$\varinjlim_{R_m\rightarrow\infty}\ \dfrac{R_1V_s}{R_1+R_2+R_m}\$

But I'm kind of confused.

So how should I do it?

  • 2
    \$\begingroup\$ This is more of a maths question .. also, as written, with Rm on the bottom only, it tends to zero? \$\endgroup\$
    – pjc50
    Mar 20 '13 at 15:44
  • 1
    \$\begingroup\$ Maybe Rm is meant to be "in parallel" with R1 - that would make sense. Then the limit will be R1/(R1+R2). Please check how you've written the formula \$\endgroup\$
    – Andy aka
    Mar 20 '13 at 15:47
  • \$\begingroup\$ @pjc50 : That's what I found at first as well since my initial approach was wrong \$\endgroup\$ Mar 20 '13 at 16:37
  • 1
    \$\begingroup\$ I've down-voted the question because it didn't show any research effort - easy research really - just read the comment i made above. \$\endgroup\$
    – Andy aka
    Mar 20 '13 at 17:27

As Andy aka points out, it's not entirely clear what you're asking. A schematic would help in the future. I think your original equation is incorrect. It sounds like you have this circuit:


simulate this circuit – Schematic created using CircuitLab

From the basic voltage divider equation: $$ V_{out}=\frac{R_LV_s}{R_L+R_2} $$

Where RL is R1 || RM.

$$ R_L=\frac{1}{\frac{1}{R_1}+\frac{1}{R_M}} $$

Showing that RL -> R1 as RM -> infinity is simple algebra from this point.

Now, as a matter of reality, RM is not infinite. This approximation only holds value if RM is much larger than R1. If RM is within, say, two orders of magnitude of the value of R1, the meter can affect the circuit it's attached to.

  • \$\begingroup\$ @Can - why did you dismiss what I said (2nd comment on your original question)? \$\endgroup\$
    – Andy aka
    Mar 20 '13 at 17:20
  • \$\begingroup\$ @Remiel - were you unable to read my comment on Can's original question? \$\endgroup\$
    – Andy aka
    Mar 20 '13 at 17:22
  • \$\begingroup\$ @Andyaka I don't disagree with your comment. The question itself is poorly phrased. But the answer, and further the fundamental concept, are still valuable. \$\endgroup\$ Mar 20 '13 at 18:23
  • \$\begingroup\$ @Remiel - would it be courtious to add to your answer that you had (or had not) seen my comment before writing your answer? \$\endgroup\$
    – Andy aka
    Mar 20 '13 at 18:38
  • \$\begingroup\$ @Andyaka Sure, if you like. \$\endgroup\$ Mar 20 '13 at 18:46

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