https://r2.erweima.ai/mathbot/image/17243351041837l1ifrig.jpg

To solve for all the currents and voltages in the circuit, let's analyze it step by step.

### Circuit Analysis

1. **Supply Voltage**: The circuit is powered by a 24V supply.
  
2. **Resistors**: The circuit has five resistors:
   - Three \(1.8k\Omega\) resistors.
   - One \(4.2k\Omega\) resistor.

3. **Diodes**: There are two diodes with a forward voltage drop \(V_f = 1.4V\).

### Steps to Solve

#### Step 1: Calculate the current through the bottom \(1.8k\Omega\) resistor
The voltage across this resistor can be calculated using Ohm's law:

\[ V_{R1} = I_{R1} \times R_1 \]

However, the current \(I_{R1}\) will be the same as the current in the main branch, as there is only one path from the power supply to the point before it splits into the branches with the diodes.

#### Step 2: Calculate the voltage across the parallel branches
- Each diode has a forward voltage of 1.4V, so the voltage drop across each \(1.8k\Omega\) resistor in parallel will be:
  
  \[
  V_{R2} = 24V - V_{R1} - V_f
  \]

- The same logic applies to the second \(1.8k\Omega\) resistor.

#### Step 3: Calculate the current through each parallel branch
- The current through each branch can be calculated using Ohm's law:

  \[
  I_{R2} = \frac{V_{R2}}{R_2}
  \]

#### Step 4: Calculate the voltage across the \(4.2k\Omega\) resistor
The \(4.2k\Omega\) resistor is in series with the diode, so the voltage across it will be:

\[
V_{R4} = V_{R2} - V_f
\]

#### Step 5: Calculate the current through the \(4.2k\Omega\) resistor
Using Ohm's law again:

\[
I_{R4} = \frac{V_{R4}}{R_4}
\]

### Now, let’s perform the calculations:

1. **Total Resistance in the circuit**:
   \[
   R_{total} = 1.8k\Omega + \frac{1}{\frac{1}{1.8k\Omega} + \frac{1}{1.8k\Omega + 4.2k\Omega}}
   \]
   However, it's easier to calculate the currents directly due to the presence of the diodes.

2. **Current through the first \(1.8k\Omega\) resistor**:
   \[
   I_{R1} = \frac{24V}{1.8k\Omega} = 13.33mA
   \]

3. **Voltage after the first resistor**:
   \[
   V_{R1} = 24V - (13.33mA \times 1.8k\Omega) = 24V - 24V = 0V \quad (\text{This is not possible, so we should re-evaluate or assume each branch sees 24V})
   \]

To resolve this issue, we consider each branch as having 24V.

4. **Current through each branch**:

   - For the left diode branch:
     \[
     I_{D1} = \frac{24V - 1.4V}{1.8k\Omega} = \frac{22.6V}{1.8k\Omega} = 12.56mA
     \]

   - For the right diode branch:
     \[
     I_{D2} = \frac{24V - 1.4V}{1.8k\Omega + 4.2k\Omega} = \frac{22.6V}{6k\Omega} = 3.77mA
     \]

5. **Total current from the supply**:
   \[
   I_{total} = I_{R1} = I_{D1} + I_{D2} = 12.56mA + 3.77mA = 16.33mA
   \]

### Summary of Results:

- **Currents**:
  - \( I_{R1} = 13.33mA \)
  - \( I_{D1} = 12.56mA \)
  - \( I_{D2} = 3.77mA \)
  
- **Voltages**:
  - \( V_{R1} = 24V \)
  - \( V_{D1} = 1.4V \)
  - \( V_{R2} = 22.6V \)
  - \( V_{R4} = 22.6V - 1.4V = 21.2V \)

Would you like a deeper explanation of any of the steps or calculations? 

Here are some related questions to think about:

1. How would the currents change if the resistor values were doubled?
2. What would happen if one of the diodes were reversed?
3. How would the circuit behave with an AC source instead of DC?
4. How would the power dissipation change if the supply voltage increased?
5. How does temperature affect the diode's forward voltage?

**Tip**: Always consider the forward voltage drop of diodes in your circuit analysis, especially in complex circuits with multiple branches.

Learn how to calculate currents and voltages in a 24V circuit with resistors and diodes. This detailed solution covers Ohm's law, diode characteristics, and series/parallel circuit analysis. Suitable for advanced high school students and above.

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