Consider the two circuits $P$ and $Q$ shown below,which are used to measure the unknown resistance $R$. In each case,the resistance is estimated by using Ohm's law $R_{\text{est}} = \frac{V}{I}$,where $V$ and $I$ are the readings of the voltmeter and the ammeter,respectively. The meter resistances $R_V$ and $R_A$ are such that $R_A \ll R \ll R_V$. The internal resistance of the battery may be ignored. The absolute error in the estimate of the resistance is denoted by $\delta R = |R - R_{\text{est}}|$.
$(a)$ Express $\delta R_P$ in terms of the given resistance values.
$(b)$ Express $\delta R_Q$ in terms of the given resistance values.
$(c)$ For what value of $R$ will $\delta R_P \approx \delta R_Q$?

(D) In circuit $P$,the voltmeter is in parallel with $R$. The measured voltage $V$ is the voltage across $R$,but the ammeter measures the total current $I = I_R + I_V = \frac{V}{R} + \frac{V}{R_V}$.
Thus,$R_{\text{est}} = \frac{V}{I} = \frac{V}{V/R + V/R_V} = \frac{R R_V}{R + R_V} = R \left(1 + \frac{R}{R_V}\right)^{-1} \approx R \left(1 - \frac{R}{R_V}\right)$.
The error is $\delta R_P = |R - R_{\text{est}}| = |R - R(1 - R/R_V)| = \frac{R^2}{R_V}$.
$(b)$ In circuit $Q$,the ammeter is in series with $R$. The ammeter measures the current $I$ through $R$,but the voltmeter measures the voltage across both $R$ and the ammeter,$V = I(R + R_A)$.
Thus,$R_{\text{est}} = \frac{V}{I} = R + R_A$.
The error is $\delta R_Q = |R - R_{\text{est}}| = |R - (R + R_A)| = R_A$.
$(c)$ For $\delta R_P \approx \delta R_Q$,we have $\frac{R^2}{R_V} = R_A$,which implies $R^2 = R_A R_V$,or $R = \sqrt{R_A R_V}$.