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To solve for all the currents and voltages in the given circuit, we can follow these steps:

### 1. **Identify the components and configuration:**
   - **Power Supply (V\_s)**: 24V
   - **Resistors**:
     - \( R_1 = 1.8 \, \text{k}\Omega \)
     - \( R_2 = 1.8 \, \text{k}\Omega \)
     - \( R_3 = 4.2 \, \text{k}\Omega \)
     - \( R_4 = 1.8 \, \text{k}\Omega \)
   - **Diodes**:
     - \( D_1 \) and \( D_2 \) with a forward voltage \( V_f = 1.4 \, \text{V} \)

### 2. **Analyze the circuit:**
   - The diodes \( D_1 \) and \( D_2 \) have a forward voltage drop of 1.4V. This means the voltage drop across each diode when it is forward biased is 1.4V.
   - \( R_1 \) and \( R_2 \) are in series with \( D_1 \) and \( D_2 \) respectively.
   - \( R_3 \) is in parallel with the diode-resistor branches.
   - \( R_4 \) is in series with the entire circuit.

### 3. **Calculate the currents through each branch:**
   - **Voltage at the top of \( R_1 \) and \( R_2 \)** (Point A): Since the diodes have a voltage drop of 1.4V, the voltage across \( R_1 \) and \( R_2 \) will be the remaining voltage after subtracting the forward voltage of the diodes.

     Let's denote the current through \( R_1 \) as \( I_1 \) and through \( R_2 \) as \( I_2 \).

     For \( R_1 \):
     \[
     I_1 = \frac{V_A - V_f}{R_1} = \frac{V_A - 1.4V}{1.8 \, \text{k}\Omega}
     \]
     For \( R_2 \):
     \[
     I_2 = \frac{V_A - V_f}{R_2} = \frac{V_A - 1.4V}{1.8 \, \text{k}\Omega}
     \]
     Where \( V_A \) is the voltage at the junction where \( R_1 \), \( R_2 \), and \( R_3 \) connect.

   - The current through \( R_3 \), \( I_3 \), can be calculated using Ohm's law:
     \[
     I_3 = \frac{V_A}{R_3} = \frac{V_A}{4.2 \, \text{k}\Omega}
     \]

   - The current through \( R_4 \), \( I_4 \), is the total current supplied by the voltage source:
     \[
     I_4 = \frac{24V - V_A}{R_4} = \frac{24V - V_A}{1.8 \, \text{k}\Omega}
     \]

### 4. **Kirchhoff’s Current Law (KCL) at point A:**
   The sum of the currents entering point A should equal the sum of the currents leaving point A.
   \[
   I_4 = I_1 + I_2 + I_3
   \]
   Substituting the expressions for \( I_1 \), \( I_2 \), and \( I_3 \):
   \[
   \frac{24V - V_A}{1.8 \, \text{k}\Omega} = \frac{V_A - 1.4V}{1.8 \, \text{k}\Omega} + \frac{V_A - 1.4V}{1.8 \, \text{k}\Omega} + \frac{V_A}{4.2 \, \text{k}\Omega}
   \]

### 5. **Solve the equation:**
   We can solve this equation to find the value of \( V_A \) and then use it to find the currents \( I_1 \), \( I_2 \), \( I_3 \), and \( I_4 \).

Let's calculate this step by step.The voltage at point A (\( V_A \)) is approximately **7.82V**.

### 6. **Calculate the currents \( I_1 \), \( I_2 \), \( I_3 \), and \( I_4 \):**
Now that we have \( V_A \), we can calculate the currents through each resistor.

Let's calculate these currents.The currents in the circuit are approximately:

- **\( I_1 \) (Current through \( R_1 \))**: 3.56 mA
- **\( I_2 \) (Current through \( R_2 \))**: 3.56 mA
- **\( I_3 \) (Current through \( R_3 \))**: 1.86 mA
- **\( I_4 \) (Current through \( R_4 \))**: 8.99 mA

### Summary of Voltages and Currents:
- Voltage at point \( A \) (\( V_A \)): **7.82V**
- Current through \( R_1 \) and \( R_2 \): **3.56 mA** each
- Current through \( R_3 \): **1.86 mA**
- Current through \( R_4 \): **8.99 mA**

Would you like further details or explanations on any part of this solution? 

### Questions for Further Exploration:
1. How would the circuit behavior change if the diode forward voltage were different?
2. What would happen if \( R_3 \) had a different resistance value?
3. Can you calculate the power dissipated by each resistor?
4. What is the power supplied by the voltage source?
5. How would the circuit be affected if the diodes were reversed?

### Tip:
When analyzing circuits with diodes, always remember to consider the forward voltage drop, as it significantly impacts the voltages and currents in the circuit.

Learn how to calculate all currents and voltages in an electrical circuit with detailed steps. This problem involves analyzing circuit components such as resistors and diodes using Ohm's Law and Kirchhoff's Current Law. Suitable for college-level students and electrical engineering enthusiasts.

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