1. **Statement of the problem:** We have an ideal electromotive force (emf) source of $12\,V$ connected in a circuit with resistors and points A, B, and C. We need to find the potential differences: a) $U_{AB} = V_A - V_B$ b) $U_{AC} = V_A - V_C$ 2. **Understanding the circuit:** The emf $\varepsilon = 12\,V$ is connected between points A and B. Since the source is ideal, it maintains a fixed voltage of 12 V between A and B regardless of current. Therefore, directly from the source: $$ U_{AB} = V_A - V_B = 12\,V $$ 3. **Finding $U_{AC}$:** Between points A and C, the circuit has three 10 $\Omega$ resistors: one directly from A to C, and two others forming a series path from A to B to C. Let's find the equivalent resistance between A and C. - The resistor from A to C: $R_1 = 10\,\Omega$ - The series path A-B-C consists of two resistors each $10\,\Omega$ (A-B and B-C): total $R_2 = 10 + 10 = 20\,\Omega$ These two paths from A to C are in parallel: $$ R_{AC} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{10 \times 20}{10 + 20} = \frac{200}{30} = \frac{20}{3} \approx 6.67\,\Omega $$ 4. **Determine voltage at point C:** Since the emf directly sets $U_{AB} = 12\,V$ and the resistors between A, B, and C, we analyze the node voltages. - Let ground be at the bottom node connected to the resistor from B to ground (10 $\Omega$). - Voltage at A is $V_A = 12\,V$ (positive terminal of the emf). - Voltage at B is $V_B = 0\,V$ (negative terminal of the emf and connected to ground through 10 $\Omega$ resistor). Using voltage division and the resistors from B to ground, and considering the bridge symmetry, the voltage at C due to the voltage divider formed by resistors is calculated as follows: Because B is at 0 V, and the resistor from B to ground is 10 $\Omega$, no current causes a voltage drop there (assuming ideal ground). Considering the parallel paths from A to C, the voltage at C is given by the voltage divider: $$ V_C = V_A - I_{AC} \times R_1 $$ Where $I_{AC}$ is the current through the resistor from A to C. We find $I_{AC}$ using total current from A to C: The total resistance between A and C is $6.67\,\Omega$, so the current from A to C is: $$ I_{AC} = \frac{V_A - V_C}{R_{AC}} $$ Since $V_C$ is unknown, rewrite in terms of currents through each path. Alternatively, by symmetry and circuit analysis, the potential at C is one-third of the way from A to B (since resistors form a voltage divider with ratios 10, 10, and 10), so: $$ U_{AC} = V_A - V_C = 8\,V $$ (This is because the 3 resistors divide the voltage 12 V proportionally, and voltage at C is $4\,V$ below $V_A$.) 5. **Final answers:** - a) $U_{AB} = 12\,V$ - b) $U_{AC} = 8\,V$ These results show the potential difference imposed directly by the emf and the drop across resistors from A to C.