Combo Circuit

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?

 nfoland Posts: 5, Reputation: 10 Junior Member #2 Feb 10, 2012, 01:12 PM
It didnt 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.
 ebaines Posts: 10,039, Reputation: 5534 Expert #3 Feb 10, 2012, 01:33 PM
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.
Attached Images

 nfoland Posts: 5, Reputation: 10 Junior Member #4 Feb 11, 2012, 12:17 AM
Thank you so much!

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