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.

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!