8. In the circuit sketched in the figure, the emf of the battery is $\mathcal{E} = 12 V$, with an internal resistance $r = 2.0 \Omega$. If the load resistor is $R = 6.0 \Omega$, then Calculate:
A. The terminal voltage of the battery. (1 point)
B. The power that is converted by the battery. (1 point)
C. The power dissipated in the internal resistance of the battery. (1 point)
D. The net power output of the battery. Compare it with the power input delivered to the load resistor. (1 point)