# Power loss in a circuit help

1. Apr 29, 2005

### poopoo16

ok consider the circuit. what is the power loss in the 20 ohm resistor in J/s?

ok so what i did was that.... and i dont know where to go after... please help

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2. Apr 29, 2005

### OlderDan

That's a good start. Now what about the 6 ohm resistor?

3. Apr 29, 2005

### whozum

Reduce the circuit to an equivalent resistor, find the flowing current, and from there use your power equations to find the power.

You are on the right track.

4. Apr 29, 2005

### poopoo16

that s my question... dont know how to treat the 6 ohms... do i add it to the 4?

and then use P=IV where I=V/R?...if i do that, its give me 1...and that is not the right answer...

5. Apr 29, 2005

### SpaceTiger

Staff Emeritus
Yes, because it's in series with the other two. Be careful, however, in the power calculation. Is the current flowing through the 20 ohm resistor the same as through the whole circuit?

6. Apr 29, 2005

### poopoo16

well, after u combine the 2 parallel ones, its no longer going thru the 20 per se right, its going thru a 6 and a 4?...soo... it then a series right?

7. Apr 29, 2005

### poopoo16

so...

R=10
V=10
I=V/R
hence I=10/10?
meanning
P=IV
P=(1)10...?

8. Apr 29, 2005

### OlderDan

What you have found IS the right answer for the current flowing through the power source and through the 6 ohm resistor. The next step is to figure out the voltage across the 20 ohm and 5 ohm resistors (both the same). You can do that by using the total current and the equivalent resistance of that parallel combination that you found in the first step, or you can calculate the voltage drop across the 6 ohm resistor when that 1 amp of current flows through it and subtract that from the 10 volt source voltage. Once you have that voltage, you can use a power equation in terms of V and R, or you can use it to calculate the current through each of those resistors and use the power equation in terms of I and V. You should note that the 1 amp of current gets divided into two parts that flow through the two paths (5 ohm and 20 ohm) and that the sum of those two parts must add up to the current flowing through the 6 ohm and the power source, the 1 amp you have already found.

9. Apr 29, 2005

### poopoo16

how do i calcuate voltyage drop arcross the 6ohm? do i use P=I^2R?

10. Apr 29, 2005

### OlderDan

You used V = IR solved for I to find the current of 1amp. Now you know the current through the 6 ohm resistor. Use V = IR to find the voltage across that one resistor.