electric-circuitselectrical-resistancehomework-and-exercisespower
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.
The 3 A that leaves the battery depends on two things: the amount of supply voltage, and the impedance (here resistance) that draws this current.
So the second part simply says:
Here is a schematic of the situation with the governing equations:
You aren't applying your power equation correctly. You can't just plug in any voltage, current, or resistance.
The equation $P=IV$ gives you the power delivered or dissipated by a single part of the circuit with a potential drop $V$ across it and current $I$ flowing through it.
If you want the total power dissipated in the circuit, you must apply this to each element separately. If just the outer resistance, then only consider this resistor. I will leave this to you.
Best Answer
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.
$$\frac{V^2}{3} = 4 \cdot \frac{V^2}{12} = 4P $$