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To solve the problem, we need to apply Kirchhoff's loop rule (KVL) and Ohm's law. The circuit includes resistors \( R_1 \), \( R_2 \), and three EMF sources \( \mathcal{E}_1 \), \( \mathcal{E}_2 \), and \( \mathcal{E}_3 \).

Given:
- \( R_1 = 140 \, \Omega \)
- \( R_2 = 69 \, \Omega \)
- \( \mathcal{E}_1 = 7.0 \, \text{V} \)
- \( \mathcal{E}_2 = 4.0 \, \text{V} \)
- \( \mathcal{E}_3 = 5.0 \, \text{V} \)

### Step-by-step plan:

1. **Label currents**:
   - Let \( I_1 \) be the current through \( R_1 \) (clockwise through the loop with \( \mathcal{E}_2 \) and \( R_1 \)).
   - Let \( I_2 \) be the current through \( R_2 \) (clockwise through the loop with \( \mathcal{E}_1 \) and \( \mathcal{E}_3 \)).

2. **Apply Kirchhoff's Voltage Law (KVL)** to each loop:
   - **Loop 1** (left loop with \( \mathcal{E}_2 \) and \( R_1 \)):
     \[
     \mathcal{E}_2 - I_1 R_1 = 0 \quad \Rightarrow \quad I_1 = \frac{\mathcal{E}_2}{R_1}
     \]
   - **Loop 2** (right loop with \( \mathcal{E}_1 \), \( \mathcal{E}_3 \), and \( R_2 \)):
     \[
     \mathcal{E}_1 - I_2 R_2 - \mathcal{E}_3 = 0 \quad \Rightarrow \quad I_2 = \frac{\mathcal{E}_1 - \mathcal{E}_3}{R_2}
     \]

3. **Solve for currents**:
   - For \( I_1 \):
     \[
     I_1 = \frac{4.0 \, \text{V}}{140 \, \Omega} = 0.02857 \, \text{A} \quad \text{(28.57 mA)}
     \]
   - For \( I_2 \):
     \[
     I_2 = \frac{7.0 \, \text{V} - 5.0 \, \text{V}}{69 \, \Omega} = \frac{2.0 \, \text{V}}{69 \, \Omega} = 0.02899 \, \text{A} \quad \text{(28.99 mA)}
     \]

4. **Calculate the potential difference between points \( a \) and \( b \)**:
   The potential difference across \( R_2 \) is:
   \[
   V_{ab} = I_2 R_2 = 0.02899 \, \text{A} \times 69 \, \Omega = 2.0 \, \text{V}
   \]

### Summary:
- **(a)** Current in \( R_1 \) is \( I_1 = 0.02857 \, \text{A} \).
- **(b)** Current in \( R_2 \) is \( I_2 = 0.02899 \, \text{A} \).
- **(c)** Potential difference between \( a \) and \( b \) is \( 2.0 \, \text{V} \).

Would you like more details or have any further questions?

### Related Questions:
1. How would the results change if the resistance of \( R_1 \) was increased?
2. What happens if the polarity of \( \mathcal{E}_1 \) is reversed?
3. How would you calculate the power dissipated in \( R_2 \)?
4. Can you apply Kirchhoff's Current Law (KCL) to this circuit, and how would it help?
5. What if one of the EMF sources had internal resistance?

### Tip:
Always pay attention to the direction of current and polarity of the voltage sources when applying Kirchhoff's laws.

In the figure R1 = 140Ω, R2 = 69Ω, and the ideal batteries have emfs E1 = 7.0 V, E2 = 4.0 V, and E3 = 5.0 V. Find (a) the current in R1, (b) the current in R2, and (c) the potential difference between points a and b.

Solve the problem of calculating the current in resistors R1 and R2 and the potential difference between points a and b using Kirchhoff's Voltage Law and Ohm's Law. This problem involves resistors R1 = 140Ω, R2 = 69Ω, and EMFs E1 = 7.0 V, E2 = 4.0 V, E3 = 5.0 V.

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