Consider a circuit in which a resistor of resistance R is attached to a battery of voltage V. A current I flows through the resistor, in accordance to Ohm's Law. Since each charged particle loses a potential energy of qV, where q is the charge of the particle, as it goes from one terminal to the other, the battery must do qV of work to pump the charge back to the initial terminal. This means that the power output of the battery is given by P = V I. If the voltage of the battery is doubled, how does the output power of the battery change? Group of answer choices The power increases by a factor of four. The power decreases by a factor of two. The power increases by a factor of two. The power does not change