I-------I------------I
I... I... I
9V... 680ohm... 330ohms
I... I... I
I... I... 150ohms
I-------I------------I

The I's, dashes, and values represent a circuit. The periods are there to fill in the space. What is the voltage, current, and power of each resistor?

It didn't make the image like I wanted it to. The 150 ohm and 330 ohm resistors are suppose to be in the same line with the 330 on top of the 150.

I redrew the circuit to make it clearer. First start with the 680 ohm resistor. Since it's connected directly to the battery, the voltage across it is 9 volts. You can find current from Ohm's law (V=iR) and power from either P=V^2/R or P = i^2R.

For the resistors on the right recall that two resistors in series are like one resistor of value R1 + R2. That single equivalent resistor is connected directly to 9 volts, so the current flow from Ohm's Law is


i = \frac {9\ volts} {(R1 + R2)}
.

Now for power - there are two ways to do this:
a) since you now know the current flow through the two resistors, you can use
P_1 = i^2 R_1 and P_2 = i^2 R_2

b) or you could recognize that the voltage drops across resistor 1 and 2 are:


V_1 = 9\ volts \ \times \ \frac {R_1}{(R_1+R_)} \\
V_2 = 9 \ volts \ \times \ \frac {R_2}{(R_1+R_2)}


Then use :


P_1 = \frac {V_1^2}{ R_1} and {P_2} = \frac {V_2^2} {R_2}.

I suggest practicing both methods so you are comfortable with them.

Thank you so much!