What is the Power Loss in a 20 Ohm Resistor in a Circuit?

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Homework Help Overview

The discussion revolves around calculating the power loss in a 20 ohm resistor within a circuit that includes other resistors, specifically a 6 ohm and a 4 ohm resistor. Participants are exploring the relationships between these components and how to approach the problem of power loss.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss reducing the circuit to an equivalent resistance and finding the current flowing through the resistors. Questions arise about how to treat the 6 ohm resistor in relation to the others and whether to add resistances in series or parallel.

Discussion Status

Some participants have provided guidance on combining resistors and calculating current, while others express uncertainty about the correct approach to finding voltage drops and power loss. Multiple interpretations of the circuit configuration are being explored.

Contextual Notes

There is a mention of a 10 volt source and the need to consider how current divides among the resistors. Participants are also questioning the assumptions regarding the current flowing through the different resistors.

poopoo16
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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 don't know where to go after... please help
 

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poopoo16 said:
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 don't know where to go after... please help

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


that s my question... don't 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...
 
poopoo16 said:
that s my question... don't know how to treat the 6 ohms... do i add it to the 4?

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?
 
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?
 
so...

R=10
V=10
I=V/R
hence I=10/10?
meanning
P=IV
P=(1)10...?
 
poopoo16 said:
that s my question... don't 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...

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.
 
how do i calcuate voltyage drop arcross the 6ohm? do i use P=I^2R?
 
  • #10
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.
 

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