Source: High school physics (Chinese)
Problem
In the circuit shown, the source has EMF $\varepsilon = 3$ V and internal resistance $r = 0.6\ \Omega$. The resistors satisfy $R_1 = R_2 = R_3 = 4\ \Omega$ and $R_4 = R_5 = 2\ \Omega$. Between the two terminal rails of the source, two horizontal branches are connected in parallel: the upper branch contains $R_4$ and $R_5$ in series, and the lower branch contains $R_1$, $R_2$, and $R_3$ in series. An ideal ammeter (A) is in the main line in series with the source, and an ideal voltmeter (V) is connected across the parallel combination (i.e., across the upper $R_4$–$R_5$ branch).
Ammeter $I \approx 0.83$ A $\bigl(\!=\tfrac{5}{6}$ A$\bigr)$; Voltmeter $U_V = 2.5$ V
\textbf{Equivalent external resistance.} The two branches in parallel:
$$R_{\text{upper}} = R_4 + R_5 = 4\ \Omega, \qquad R_{\text{lower}} = R_1 + R_2 + R_3 = 12\ \Omega,$$ $$R_{\text{ext}} = \frac{R_{\text{upper}} R_{\text{lower}}}{R_{\text{upper}} + R_{\text{lower}}} = \frac{4 \times 12}{4 + 12} = 3\ \Omega.$$\textbf{Total current (ammeter reading).}
$$I = \frac{\varepsilon}{R_{\text{ext}} + r} = \frac{3}{3 + 0.6} = \frac{5}{6}\ \text{A} \approx 0.83\ \text{A}.$$\textbf{Voltmeter reading} (voltage across the parallel section):
$$U_V = I \, R_{\text{ext}} = \frac{5}{6} \times 3 = 2.5\ \text{V}.$$(Equivalent check: $U_V = \varepsilon - I r = 3 - \tfrac{5}{6} \times 0.6 = 3 - 0.5 = 2.5$ V.)