1. **Stating the problem:** We have a rectangular circuit with three ammeters measuring currents $i_1$, $i_2$, and $i_3$. The total current entering the circuit is 250 mA. We need to find the values of $i_1$, $i_2$, and $i_3$. 2. **Understanding the circuit and applying Kirchhoff's Current Law (KCL):** KCL states that the total current entering a junction equals the total current leaving it. Here, the currents $i_1$, $i_2$, and $i_3$ are related by the current flow in the circuit. 3. **Expressing the total current:** The total current $I_{total} = 250$ mA is the sum of the currents through the ammeters. Assuming the currents $i_1$, $i_2$, and $i_3$ flow through different branches, we have: $$i_1 + i_2 + i_3 = 250 \text{ mA}$$ 4. **Using the direction of currents and circuit symmetry:** From the figure description, $i_r$ flows downwards between the batteries, $i_1$ flows upwards on the right vertical segment, $i_2$ flows left to right on the bottom horizontal segment, and $i_3$ flows upwards on the left vertical segment. 5. **Applying Kirchhoff's Voltage Law (KVL) or additional circuit rules:** Without resistance or voltage values, we assume the currents are equal due to symmetry or given conditions. Thus: $$i_1 = i_2 = i_3$$ 6. **Solving for each current:** Using the total current equation: $$3i_1 = 250 \Rightarrow i_1 = \frac{250}{3} \approx 83.33 \text{ mA}$$ Therefore: $$i_1 = i_2 = i_3 \approx 83.33 \text{ mA}$$ **Final answer:** The currents indicated by the three ammeters are approximately 83.33 mA each.