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.
Best Answer
Model the circuit without the voltage follower as so:
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}}\$.