Any help is appreciated(: - A circuit has 12 Ω of resistance and is connected to a 12 v battery. What is the change in power if the resistance decreases to 9 Ω?

Question

theeric · Answer

Hi! So, I think we can try the technique of finding the power with the $12\ [\Omega]$ resistor and call it something like $P_{12}$. Then we can find the power with the $9\ [\Omega]$ resistor, and call it $P_9$. Then we can compare the two and find the change in power!

You can use any two of voltage, resistance, and current to make a formula for power. So let's look at voltage and resistance. I always remember the simplest equation, $P=IV$, and then use Ohm's law if I forget the others. Since $I=\dfrac{V}{R}$, $P=\dfrac{V^2}{R}$. So now you can calculate $P_{12}$ and $P_9$.

theeric · Answer

The change in power from $P_{12}$ to $P_9$ is $\Delta P=P_9-P_{12}$, always $\Delta \text{anything}=\text{final}-\text{initial}$.

anonymous · Answer

Wouldn't that be -3? .-.

theeric · Answer

What did you get for $P_9$?

anonymous · Answer

I do not understand this at all ._.

doc.brown · Answer

Less resistance, more stuff can get through.
The initial power
P=V^2/R
P= 12*12/12
P= 12W

Change the resistor to a 9ohm
P= 12*12/9
P= 16W

The change is final minus initial.
16-12
= 4W

anonymous · Answer

Thanks so much!