Find the individual currents that flow through each of the following resistors: $$ R_{1}=3 \Omega, R_{2}=5 \Omega, R_{3}=5 \Omega, R_{4}=10 \Omega $$ $$ \left(R_{2}\right. $$ and $$ R_{3} $$ are parallel) $$ E=24 \mathrm{~V} $$

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To determine the individual currents flowing through each resistor in the given circuit, we'll follow these steps:

1. **Understand the Circuit Configuration:**
   
   - **Resistors:** $$ R_1 = 3\,\Omega $$, $$ R_2 = 5\,\Omega $$, $$ R_3 = 5\,\Omega $$, $$ R_4 = 10\,\Omega $$
   - **Voltage Source:** $$ E = 24\,\text{V} $$
   - **Configuration:** $$ R_2 $$ and $$ R_3 $$ are in parallel, and this parallel combination is in series with $$ R_1 $$ and $$ R_4 $$.

![Circuit Diagram](https://i.imgur.com/XYZabcD.png) *(Assumed Configuration)*

2. **Calculate the Equivalent Resistance:**

First, find the equivalent resistance of the parallel combination of $$ R_2 $$ and $$ R_3 $$:
   $$
   R_{23} = \frac{R_2 \times R_3}{R_2 + R_3} = \frac{5\,\Omega \times 5\,\Omega}{5\,\Omega + 5\,\Omega} = \frac{25}{10} = 2.5\,\Omega
   $$

Next, sum the resistances in series:
   $$
   R_{\text{total}} = R_1 + R_{23} + R_4 = 3\,\Omega + 2.5\,\Omega + 10\,\Omega = 15.5\,\Omega
   $$

3. **Determine the Total Current ($$ I_{\text{total}} $$):**
   $$
   I_{\text{total}} = \frac{E}{R_{\text{total}}} = \frac{24\,\text{V}}{15.5\,\Omega} \approx 1.548\,\text{A}
   $$

4. **Find the Voltage Drops Across Each Component:**
   
   - **Across $$ R_1 $$:**
     $$
     V_1 = I_{\text{total}} \times R_1 = 1.548\,\text{A} \times 3\,\Omega \approx 4.644\,\text{V}
     $$
   
   - **Across $$ R_{23} $$ (Parallel Combination of $$ R_2 $$ and $$ R_3 $$):**
     $$
     V_{23} = I_{\text{total}} \times R_{23} = 1.548\,\text{A} \times 2.5\,\Omega \approx 3.870\,\text{V}
     $$
   
   - **Across $$ R_4 $$:**
     $$
     V_4 = I_{\text{total}} \times R_4 = 1.548\,\text{A} \times 10\,\Omega \approx 15.48\,\text{V}
     $$

5. **Calculate Currents Through $$ R_2 $$ and $$ R_3 $$:**
   
   Since $$ R_2 $$ and $$ R_3 $$ are in parallel, the voltage across them is the same ($$ V_{23} $$).
   
   - **Current Through $$ R_2 $$ ($$ I_2 $$):**
     $$
     I_2 = \frac{V_{23}}{R_2} = \frac{3.870\,\text{V}}{5\,\Omega} \approx 0.774\,\text{A}
     $$
   
   - **Current Through $$ R_3 $$ ($$ I_3 $$):**
     $$
     I_3 = \frac{V_{23}}{R_3} = \frac{3.870\,\text{V}}{5\,\Omega} \approx 0.774\,\text{A}
     $$

6. **Summary of Individual Currents:**

| Resistor | Current ($$ I $$) |
   |----------|-------------------|
   | $$ R_1 $$ | $$ \approx 1.548\,\text{A} $$ |
   | $$ R_2 $$ | $$ \approx 0.774\,\text{A} $$ |
   | $$ R_3 $$ | $$ \approx 0.774\,\text{A} $$ |
   | $$ R_4 $$ | $$ \approx 1.548\,\text{A} $$ |

**Final Answer:**

- **$$ R_1 $$:** Approximately 1.548 A
- **$$ R_2 $$:** Approximately 0.774 A
- **$$ R_3 $$:** Approximately 0.774 A
- **$$ R_4 $$:** Approximately 1.548 A

All currents are calculated based on the total current of approximately 1.548 A flowing through $$ R_1 $$ and $$ R_4 $$, with $$ R_2 $$ and $$ R_3 $$ each carrying half of that total current in the parallel branch.
The currents through each resistor are: - $$ R_1 $$: 1.548 A - $$ R_2 $$: 0.774 A - $$ R_3 $$: 0.774 A - $$ R_4 $$: 1.548 A

Find the individual currents that flow through each of the following resistors: \( R_{1}=3 \Omega, R_{2}=5 \Omega, R_{3}=5 \Omega, R_{4}=10 \Omega \) \( \left(R_{2}\right. \) and \( R_{3} \) are parallel) \( E=24 \mathrm{~V} \)

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