Find the individual currents that flow through each of the following resistors: $$ R_{1}=30 \Omega, R_{2}=50 \Omega, R_{3}=20 \Omega, R_{4}=40 \Omega $$ $$ \left(R_{2}\right. $$ and $$ R_{3} $$ are parallel) $$ E=60 \mathrm{~V} $$

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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:**
     - $$ R_1 = 30\,Ω $$
     - $$ R_2 = 50\,Ω $$
     - $$ R_3 = 20\,Ω $$
     - $$ R_4 = 40\,Ω $$
   
   - **Configuration Assumption:**
     - $$ R_2 $$ and $$ R_3 $$ are connected in **parallel**.
     - This parallel combination is connected in series with $$ R_1 $$ and $$ R_4 $$.
     - The entire arrangement is connected to a voltage source $$ E = 60\,V $$.

The circuit can be visualized as:
   $$
   E \rightarrow R_1 \rightarrow (R_2 \parallel R_3) \rightarrow R_4 \rightarrow \text{Return to } E
   $$

2. **Calculate Equivalent Resistance:**
   
   - **Parallel Combination of $$ R_2 $$ and $$ R_3 $$:**
     $$
     R_{\text{parallel}} = \frac{R_2 \times R_3}{R_2 + R_3} = \frac{50 \times 20}{50 + 20} = \frac{1000}{70} \approx 14.29\,Ω
     $$
   
   - **Total Resistance in the Circuit:**
     $$
     R_{\text{total}} = R_1 + R_{\text{parallel}} + R_4 = 30 + 14.29 + 40 = 84.29\,Ω
     $$

3. **Determine the Total Current ($$ I_{\text{total}} $$) from the Source:**
   $$
   I_{\text{total}} = \frac{E}{R_{\text{total}}} = \frac{60}{84.29} \approx 0.711\,A
   $$

4. **Calculate Voltage Drops Across Each Resistor:**
   
   - **Voltage Across $$ R_1 $$:**
     $$
     V_1 = I_{\text{total}} \times R_1 = 0.711 \times 30 \approx 21.43\,V
     $$
   
   - **Voltage Across the Parallel Combination ($$ R_2 $$ and $$ R_3 $$):**
     $$
     V_{\text{parallel}} = I_{\text{total}} \times R_{\text{parallel}} = 0.711 \times 14.29 \approx 10.14\,V
     $$
   
   - **Voltage Across $$ R_4 $$:**
     $$
     V_4 = I_{\text{total}} \times R_4 = 0.711 \times 40 \approx 28.57\,V
     $$

*(Verification: $$ V_1 + V_{\text{parallel}} + V_4 \approx 60\,V $$)*

5. **Determine Individual Currents Through Each Resistor:**
   
   - **Current Through $$ R_2 $$:**
     $$
     I_2 = \frac{V_{\text{parallel}}}{R_2} = \frac{10.14}{50} \approx 0.203\,A
     $$
   
   - **Current Through $$ R_3 $$:**
     $$
     I_3 = \frac{V_{\text{parallel}}}{R_3} = \frac{10.14}{20} \approx 0.507\,A
     $$

- **Current Through $$ R_1 $$ and $$ R_4 $$:**
     $$
     I_1 = I_4 = I_{\text{total}} \approx 0.711\,A
     $$

6. **Summary of Individual Currents:**

- **$$ I_1 $$ (Through $$ R_1 $$):** $$ \approx 0.71\,A $$
   - **$$ I_2 $$ (Through $$ R_2 $$):** $$ \approx 0.20\,A $$
   - **$$ I_3 $$ (Through $$ R_3 $$):** $$ \approx 0.51\,A $$
   - **$$ I_4 $$ (Through $$ R_4 $$):** $$ \approx 0.71\,A $$

**Final Answer:**
- Current through R₁: 0.71 A
- Current through R₂: 0.20 A
- Current through R₃: 0.51 A
- Current through R₄: 0.71 A
The currents through each resistor are: - $$ R_1 $$: 0.71 A - $$ R_2 $$: 0.20 A - $$ R_3 $$: 0.51 A - $$ R_4 $$: 0.71 A

Find the individual currents that flow through each of the following resistors: \( R_{1}=30 \Omega, R_{2}=50 \Omega, R_{3}=20 \Omega, R_{4}=40 \Omega \) \( \left(R_{2}\right. \) and \( R_{3} \) are parallel) \( E=60 \mathrm{~V} \)

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