Combo Circuit

nfoland · 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 · 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.

nfoland · Feb 11, 2012, 12:17 AM

Thank you so much!

Combo Circuit

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