At what rate is thermal energy being generated in the 2R-resistor when ε = 12 V and R = 3.0 Ω?

Answer:
6 W
Explanation:
From the given information:
The resistance in Parallel for 2R is:
[tex]R_p = \dfrac{2R\times 2R}{2R+2R} \\ \\ R_p= R[/tex]
The equivalent resistance:
[tex]R_{eq} = R_p + R = 2R[/tex]
[tex]R_{eq} = 2(3)[/tex]
[tex]R_{eq} = 6 \ \ ohms[/tex]
The current through the circuit in R is:
[tex]= \dfrac{12}{R+R} \\ \\ = \dfrac{12}{2\times 3} \\ \\ = 2 A[/tex]
The current through the circuit in 2R is:
[tex]I_2R = (2A) \times \dfrac{2R}{2R+2R}[/tex]
[tex]I_2R = 2A \times \dfrac{1}{2} \\ \\ I_2R = 1A[/tex]
Finally, the thermal energy:
[tex]P_{2R} = (1)^2 (2R)[/tex]
[tex]P_{2R} = (1)^2 (2\times 3)[/tex]
[tex]P_{2R} = 6W[/tex]