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The problem involves solving for the currents in a given electrical circuit, specifically using Kirchhoff's Voltage Law (KVL) and Kirchhoff's Current Law (KCL).

Let's go step by step.

### Given Values:
- \( E_1 = 10 \, V \)
- \( E_2 = 15 \, V \)
- \( R_1 = 100 \, \Omega \)
- \( R_2 = 200 \, \Omega \)
- \( R_3 = 100 \, \Omega \)
- \( R_4 = 300 \, \Omega \)
- \( R_5 = 50 \, \Omega \)

### The circuit diagram includes:
- A voltage source \( E_1 \) connected in series with resistors \( R_1 \) and \( R_3 \).
- Another voltage source \( E_2 \) connected in series with resistors \( R_2 \) and \( R_4 \).
- The resistor \( R_5 \) is connected between two points in the circuit.

### Steps to Solve:

1. **Label the Currents**: 
   - Let's denote the current in the loop with \( E_1 \) as \( I_1 \).
   - The current in the loop with \( E_2 \) as \( I_2 \).
   - The current through \( R_5 \) is \( I_3 \).

2. **Apply KCL at the Junction**: 
   - At the junction, where \( R_1 \), \( R_2 \), and \( R_5 \) meet, we have:
   \[
   I_3 = I_1 - I_2
   \]

3. **Apply KVL to the Loops**:
   - For the left loop (with \( E_1 \)):
     \[
     E_1 - I_1 R_1 - I_3 R_3 = 0
     \]
     Substituting \( I_3 = I_1 - I_2 \):
     \[
     10 - 100 I_1 - 100 (I_1 - I_2) = 0
     \]
     Simplify to:
     \[
     200 I_1 - 100 I_2 = 10
     \]
     \[
     2 I_1 - I_2 = 0.1 \quad \text{(Equation 1)}
     \]

   - For the right loop (with \( E_2 \)):
     \[
     E_2 - I_2 R_2 - I_3 R_4 = 0
     \]
     Substituting \( I_3 = I_1 - I_2 \):
     \[
     15 - 200 I_2 - 300 (I_1 - I_2) = 0
     \]
     Simplify to:
     \[
     300 I_2 - 300 I_1 = 15
     \]
     \[
     I_2 - I_1 = 0.05 \quad \text{(Equation 2)}
     \]

4. **Solve the Simultaneous Equations**:
   From Equation 2:
   \[
   I_2 = I_1 + 0.05
   \]

   Substitute into Equation 1:
   \[
   2 I_1 - (I_1 + 0.05) = 0.1
   \]
   Simplify:
   \[
   I_1 - 0.05 = 0.1
   \]
   \[
   I_1 = 0.15 \, A
   \]
   
   Substituting \( I_1 \) back into Equation 2:
   \[
   I_2 = 0.15 + 0.05 = 0.2 \, A
   \]

   Now, \( I_3 = I_1 - I_2 = 0.15 - 0.2 = -0.05 \, A \) (negative sign indicates direction opposite to assumed direction).

### Conclusion:
- \( I_1 = 0.15 \, A \)
- \( I_2 = 0.2 \, A \)
- \( I_3 = -0.05 \, A \)

These currents can be used to further analyze the voltage drops across each resistor or to find power dissipated if needed.

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Would you like further details on any step or help with a different problem?

### Follow-Up Questions:
1. What are the voltage drops across each resistor using these current values?
2. How would the analysis change if one of the resistors had a different value?
3. What if the resistors were connected in parallel instead of series?
4. How would the current distribution change if \( E_1 \) and \( E_2 \) had different polarities?
5. How do you calculate the power dissipated in each resistor?

### Tip:
Always double-check the signs of the currents and voltages, especially when using KVL and KCL, to ensure they match the assumed directions.

Learn how to solve an electric circuit problem using Kirchhoff's Voltage Law (KVL) and Kirchhoff's Current Law (KCL). This problem involves analyzing currents in a series circuit with multiple voltage sources and resistors. Detailed step-by-step solution provided.

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