Source: High school physics (Chinese)
Problem
In the circuit of Figure, the source has EMF $\varepsilon = 4$ V and internal resistance $r$. The external network has $R_1 = 4$ $\Omega$ and the series pair $R_2 + R_3 = 2 + 10 = 12$ $\Omega$ connected as two parallel branches between two nodes, and the parallel combination is in series with $R_4 = 6$ $\Omega$. An ideal ammeter is in series with the $R_1$ branch and reads $0.3$ A. Voltmeter $V_1$ is connected across $R_3$ in the middle branch; voltmeter $V_2$ is connected across the source terminals.
- Find the readings of the two voltmeters $V_1$ and $V_2$.
- Find the internal resistance $r$ of the source.
(1) $V_1 = 1$ V, $V_2 = 3.6$ V; \quad (2) $r = 1$ $\Omega$.
The two parallel branches share the same voltage $U_p$. Through the $R_1$ branch the ammeter reads $I_1 = 0.3$ A, so:
$$U_p = I_1 R_1 = 0.3 \times 4 = 1.2 \text{ V}$$Current through the $R_2$-$R_3$ branch:
$$I_2 = \frac{U_p}{R_2 + R_3} = \frac{1.2}{12} = 0.1 \text{ A}$$(1) Voltmeter readings.
$V_1$ reads the drop across $R_3$: $$V_1 = I_2 R_3 = 0.1 \times 10 = 1 \text{ V}$$The total source current is $I = I_1 + I_2 = 0.4$ A. The terminal voltage of the source equals the drop across the external load:
$$V_2 = U_p + I R_4 = 1.2 + 0.4 \times 6 = 3.6 \text{ V}$$(2) Internal resistance. From $\varepsilon = V_2 + I r$:
$$r = \frac{\varepsilon - V_2}{I} = \frac{4 - 3.6}{0.4} = 1 \text{ }\Omega$$