Finding power in series circuit

A resistor of resistance 12 ohms is connected in series with a cell of negligible internal resistance. The power dissipated in the resistor is P. The resistor is replaced with a resistor of resistance 3 ohms. What is the power dissipated in the resistor?

A) 0.25 P

B) P

C) 2 P

D) 4 P

I used $P = \frac{V^2}{R}$ and I set V = 1. So P would equal 12 W and the new P would equal 3 W. I'm confused on how the answer is 4 P and not .25 P.

You're just thinking 'upside down'.

The 3 ohm resistor is just 1/4 the resistance of the 12 ohm resistor but, power is inversely proportional to the resistance.

Thus, if you decrease the resistance by a factor of X, you increase the power by a factor of X.

In this case, the resistance is decreased by a factor of 4 so the power is increased by a factor of 4.

$$\frac{V^2}{3} = 4 \cdot \frac{V^2}{12} = 4P$$

what is your logic behind setting V=1? you do not have enough info to assume that. What you do know is that the voltage is constant (same cell), so you can write:

$P_{ini}*R_{ini}=P_{final}*R_{final}$

Knowing both R's gets you

$P_{final}=4P_{ini}$