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To solve this circuit problem, we need to find the total current (I) flowing through the circuit.

### Step-by-Step Solution:

1. **Given Data**:
   - \( R_1 = 6 \, \Omega \)
   - \( R_2 = 4 \, \Omega \)
   - Internal resistance of the battery \( r = 0.6 \, \Omega \)
   - Electromotive force (E) = 6 V

2. **Finding Equivalent Resistance**:
   - The two resistors \( R_1 \) and \( R_2 \) are in parallel, so the equivalent resistance of the parallel combination (\( R_{12} \)) is:
     \[
     \frac{1}{R_{12}} = \frac{1}{R_1} + \frac{1}{R_2}
     \]
     \[
     \frac{1}{R_{12}} = \frac{1}{6} + \frac{1}{4} = \frac{2}{12} + \frac{3}{12} = \frac{5}{12}
     \]
     Therefore, 
     \[
     R_{12} = \frac{12}{5} = 2.4 \, \Omega
     \]

3. **Total Resistance**:
   - Now, include the internal resistance of the battery, which is in series with the parallel combination. The total resistance \( R_{\text{total}} \) is:
     \[
     R_{\text{total}} = R_{12} + r = 2.4 \, \Omega + 0.6 \, \Omega = 3.0 \, \Omega
     \]

4. **Total Current**:
   - Using Ohm’s law, the total current \( I \) in the circuit is:
     \[
     I = \frac{E}{R_{\text{total}}} = \frac{6 \, \text{V}}{3.0 \, \Omega} = 2.0 \, \text{A}
     \]

Thus, the total current in the circuit is **2.0 A**.

The correct answer is **D. 2.0 A**.

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Would you like more details on any part of this explanation? Here are some related questions for further practice:

1. What would the total current be if the internal resistance of the battery was 1 Ω instead of 0.6 Ω?
2. How does the total current change if we increase the voltage to 12 V?
3. What happens to the total resistance if \( R_2 \) is doubled?
4. How would you calculate the power dissipated by \( R_1 \) in this circuit?
5. What would happen to the current if the resistors were connected in series instead of parallel?

**Tip:** When resistors are in parallel, their equivalent resistance is always less than the smallest individual resistor value!

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Learn how to calculate the total current in a circuit with parallel resistors R1 = 6 Ω and R2 = 4 Ω, and a 6 V battery with internal resistance r = 0.6 Ω using Ohm's Law and equivalent resistance formulas.

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