In the circuit depicted in the figure, the battery has an...

Based on the given circuit diagram, the switch $S$ connects the voltmeter in parallel with the external resistance $R$ only.

First, we calculate the equivalent resistance of the parallel combination of the resistor $R$ and the voltmeter $R_V$:
$R_p = \dfrac{R \times R_V}{R + R_V}$
Substituting the given values ($R = 50 \text{ }\Omega$, $R_V = 200 \text{ }\Omega$):
$R_p = \dfrac{50 \times 200}{50 + 200} = \dfrac{10000}{250} = 40 \text{ }\Omega$

This parallel combination is in series with the ammeter ($R_A = 2.0 \text{ }\Omega$) and the internal resistance of the battery ($r = 1.0 \text{ }\Omega$). The total equivalent resistance of the circuit is:
$R_{eq} = R_p + R_A + r$
$R_{eq} = 40 + 2.0 + 1.0 = 43 \text{ }\Omega$

The total current $I$ flowing through the main circuit, which is also the reading of the ammeter, is given by Ohm's law:
$I = \dfrac{\mathcal{E}}{R_{eq}}$
$I = \dfrac{4.3}{43} = 0.1 \text{ A}$

The voltmeter measures the potential difference across the parallel combination $R_p$. This voltage $V$ is:
$V = I \times R_p$
$V = 0.1 \times 40 = 4.0 \text{ V}$

Therefore, the ammeter reads $0.1 \text{ A}$ and the voltmeter reads $4.0 \text{ V}$.

Answer: $0.1 \text{ A}$, $4.0 \text{ V}$