= Loaded voltage divider
{title2=$V_{\mathrm{out}}=V_{\mathrm{in}}(R_2\parallel R_L)/(R_1+R_2\parallel R_L)$}
Connect a passive <electrical load> of <electrical resistance> $R_L>0$ from the junction to the reference. In the steady <direct current> model, $R_L$ and $R_2$ share the same <electric potential difference> and are connected in parallel. Their equivalent <electrical resistance> is
$$
R_p=R_2\parallel R_L=\frac{R_2R_L}{R_2+R_L},\qquad
V_{\mathrm{out}}=V_{\mathrm{in}}\frac{R_p}{R_1+R_p}
=\frac{V_{\mathrm{in}}R_2R_L}{R_1R_2+(R_1+R_2)R_L}.
$$
This is the replacement required when a <voltmeter> is connected across $R_2$. The inspected teaching guide explicitly models the meter this way.<Lab 4: Voltage Measurements and Meter Loading Effects, Southern Illinois University ET 304A>[ET 304A, Lab 4: Voltage Measurements and Meter Loading Effects, pp. 1–2, Figure 1 and equations (1)–(3)]{ref}
The <Kirchhoff's current law> proof makes the changed assumption visible:
$$
\frac{V_{\mathrm{in}}-V_{\mathrm{out}}}{R_1}
=\frac{V_{\mathrm{out}}}{R_2}+\frac{V_{\mathrm{out}}}{R_L}.
$$
The upper <resistor> carries the sum of the lower and load <electric currents>, so the two original <resistors> no longer have equal <electric currents>. In the stipulated $R_L=10\,\mathrm{k\Omega}$ example, $R_p=5\,\mathrm{k\Omega}$, the upper current is $2/3\,\mathrm{mA}$, and each lower branch carries $1/3\,\mathrm{mA}$.
For a digital <multimeter> used as a <voltmeter>, let its input <electrical resistance> on the selected voltage range be $R_m$. With no other load, set $R_L=R_m$; if an intended load is also present, use $R_L\parallel R_m$ in place of $R_L$. Two instruments connected simultaneously add two branches. Their ideal indicated value equals the voltage of this altered circuit; <voltmeter loading> can exist even with perfect <measurement calibration>. Real readout additionally has <measurement uncertainty>, offsets, and finite resolution. Consult the actual range and input specifications rather than assigning every digital <multimeter> the same $R_m$.
For an <oscilloscope>, the resistance-only model applies to settled <direct current> with a compatible input coupling setting. The <oscilloscope> input and cable also have <capacitance>; changing <signals> can therefore have loading that depends on <frequency>. A probe changes the impedance seen at its tip and the relation between tip voltage and displayed voltage; use the probe-and-input combination's specification and attenuation setting. NJIT's guide models the parallel <electrical resistance> and <capacitance> and adjusts probe compensation with a square-wave readout.<Lab 7: Input Impedance of an Oscilloscope and the Scope Probe, NJIT ECE 291>[Lab 7: Input Impedance of an Oscilloscope and the Scope Probe, Introduction, Figure 5 and probe adjustment; Laboratory §1(a)–(c)]{ref}
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