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In my electronics lab we constructed a voltage divider that consisted of two \$1\$M\$\Omega\$ resistors and measured the voltage drop between the two resistors and ground. We supplied \$15\$V and calculated the voltage drop via the voltage divider equation to be \$15\$V\$/2 = 7.5\$V. We measured the voltage and it was around \$5\$V. My professor explained that this erroneous reading was due to the high resistance interfering with the measurement.

We then fed the voltage from the voltage divider into an op-amp acting as a voltage follower and again measured the output voltage. We now saw the expected \$7.5\$V.

From this knowledge I must calculate the input resistance of the DMM and I'm not exactly sure where to start. Is it acceptable to view the first voltage measurement as a voltage associated with a resistance, say \$R+R_{\text{int}}\$, and the second voltage measurement associated with \$R\$, since this the voltage follower pretty much negates the input resistance of the DMM? I feel like I'm on the right track, but I'm not sure where to go next.

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Model the circuit without the voltage follower as so:

schematic

simulate this circuit – Schematic created using CircuitLab

This is still just a voltage divider, but you have \$R_2||R_{\text{DMM}}\$ as the lower resistor instead of just \$R_2\$ as in the unloaded case (and the case with the voltage follower, which ideally has infinite input impedance).

Since you measured \$5\$V at the output your voltage divider gave you the following:

$$5\text{V} = 15\text{V} \times \frac{R_2||R_{\text{DMM}}}{R_1 + R_2||R_{\text{DMM}}}$$

There is only one unknown in this equation so with some algebra you can work out \$R_{\text{DMM}}\$.

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  • \$\begingroup\$ Just one question. Why is it the above instead of \$\frac{R_2}{R_1+R_2+R_{DMM}}\$ for the fractional part of the voltage divider equation? Because the voltage across \$R_2\$ is the same as the voltage across \$R_{DMM}\$, and they are in parallel; Like how Majenko has his circuit in his solution above. \$\endgroup\$ – Lefty Nov 21 '14 at 0:46
  • \$\begingroup\$ Since \$R_2\$ and \$R_{DMM}\$ are in parallel, you can replace them with a single resistor \$R_x = R_2||R_{DMM}\$. Now you have a simple voltage divider of two resistances \$R_1\$ and \$R_x\$. The fractional part is therefore \$\frac{R_x}{R_1 + R_x}\$, which is the same as what I put once you substitute for \$R_x\$. \$\endgroup\$ – Null Nov 21 '14 at 1:30
  • \$\begingroup\$ Not a problem. If you feel that your question has been answered please accept the answer that was most useful to you (I won't be offended if you choose Majenko's). This helps the system understand that the question has been fully answered. \$\endgroup\$ – Null Nov 21 '14 at 1:47
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When you're measuring the voltage direct with the DMM you basically have this arrangement:

schematic

simulate this circuit – Schematic created using CircuitLab

From that you can calculate what the total resistance in the lower half must be in order to get 5V at the junction.

When you have calculated that total resistance, you can then separate out the two resistors in parallel to work out what R3 must equate to.

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