Resistance Measurement

Here is a problem that was recently sent to me:

To solve, I will refer to the following four configurations. The first two are the ones presented in the problem (with the switches in different state), and the third and fourth are more useful configurations.

Configuration 1:

The voltage across the voltmeter can be expressed in terms of the current through the ammeter:

$V = I \frac{R R_v}{R + R_v}$

The measured resistance in configuration 1 is therefore:

$R_{m1} =\frac{V}{I} = \frac{R R_v }{R + R_v} \\ \Delta_1 = \frac{R_v }{R + R_v} - 1 = -\frac{R}{R + R_v}$

Configuration 2:

The current through the ammeter can be expressed in terms of the voltage across the voltmeter:

$I = \frac{V}{R + R_A}$

The measured resistance in configuration 2 is therefore:

$R_{m2} = \frac{V}{I} = R + R_A \\ \Delta_2 = 1 + \frac{R_A}{R} - 1 = \frac{R_A}{R}$

There are two ways to answer part (b) of the question: using either configuration 3 or configuration 4.

Configuration 3:

Configuration 3 is a more useful form of configuration 1. Starting with the switch in position 2, we know the current through $R_v$ as well as the voltage across it, which gives us $R_v$. Then with the switch in position 1, we get $R_m = \frac{R R_v }{R + R_v}$ (which is the same result from configuration 1). And since the first measurement gave us $R_v$, we can solve for $R$.

Configuration 4:

Configuration 4 is a more useful form of configuration 2. Starting with the switch in position 1, we know the current through $R_A$ as well as the voltage across it, which gives us $R_A$. Then with the switch in position 2, we get $R_m = R + R_A$ (which is the same result from configuration 2). And since the first measurement gave us $R_A$, we can solve for $R$.