 Voltage divider loading error

For $R_1,R_2,R_L>0$ and $V_s\ne 0$, define the fractional shortfall in output magnitude from the unloaded value byThis is a loading bias relative to the unloaded target, not the uncertainty of an instrument's measurement of the connected circuit. The equality with the signed ratio also holds for negative $V_s$, because the output retains its sign. The Thévenin resistance $R_{\mathrm{th}}=R_1\parallel R_2$ summarizes the divider's susceptibility: the relevant comparison is $R_L/R_{\mathrm{th}}$, not merely a large absolute $R_L$.

Set $a=R_1/R_2>0$ and $x=R_L/R_2>0$. ThenThe derivative proves that increasing the load electrical resistance always reduces this error. For equal divider arms, $e=1/(1+2x)$: $x=1$ gives $33\frac{1}{3}\%$, $x=10$ gives $100/21\%=4.7619\dots \%$, and $x=100$ gives $100/201\%=0.497512\dots \%$. Thus a “ten times the lower resistor” rule does not guarantee one-percent agreement with the unloaded target.

A quantitative design requirement follows without a rule of thumb. For a specified $0<ϵ<1$,For equal $10\mathrm{k\Omega }$ arms, $R_{\mathrm{th}}=5\mathrm{k\Omega }$, so one-percent loading error requires $R_L\ge 495\mathrm{k\Omega }$. Reducing both divider electrical resistances reduces loading but increases unloaded supply electric current and Joule heating: $P=V_s^2/(R_1+R_2)$. The stipulated equal-arm example dissipates $5\mathrm{mW}$ in total when unloaded. This tradeoff explains why the ratio alone does not determine a useful divider.