Power in 6.0 Ohm Resistor - Circuit Related Q

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Homework Help Overview

The discussion revolves around determining the power dissipated in a 6.0 Ohm resistor within a circuit that includes other resistors and a voltage source. The problem is situated in the context of circuit analysis, particularly focusing on power calculations and current distribution.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various methods to calculate power, including the use of formulas like P = V^2/R and P = I^2R. There are questions about the differences in power across resistors and how to determine individual currents in the circuit. Some participants suggest using Kirchhoff's laws for analysis.

Discussion Status

The discussion is ongoing, with participants exploring different approaches to find the currents and power in the circuit. Some guidance has been offered regarding the use of Kirchhoff's laws, but there is no explicit consensus on the best method to proceed.

Contextual Notes

Participants express uncertainty about the values needed to calculate individual currents, particularly in the absence of specific voltage measurements for each component. There is also mention of a similar problem that may provide additional context.

MD2000
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Determine the power dissipated in the 6.0 resistor in the circuit shown in the drawing. (R1 = 4.0 , R2 = 6.0 and V1 = 15 V.)

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Am I supposed to be using P = V^2/R..is the power flowing through each resistor going to be diff?

Anyone want to help me out where to start?
 
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You can solve this many different ways. First note that the current through the 2ohm and the 1ohm resistor is the same, and that the current through R2 and the 1ohm resistor is the same.

I would first solve for the currents, and then you know that:

[tex]P=IV[/tex]

[tex]V=IR[/tex]

Thus,

[tex]P=I(IR)=I^2R[/tex]

Do you know KVL, KCL?
 
How exactly can I get the currents for each individual peice?

I know that the total current is going to be V = IR..which equals 2.46..

But how can I solve for individual currents? Since you don't have V..what can you use to get the I?
 
Last edited:
MD2000 said:
How exactly can I get the currents for each individual peice?

I know that the total current is going to be V = IR..which equals 2.46..

But how can I solve for individual currents? Since you don't have V..what can you use to get the I?
Think Kirchoff's laws, you can use these the calculate the voltage or current at each point in the circuit. :wink:
 
I have a similar question with multiple batteries and resistors in one big circuit. I have been given two labelled junctions but within one smaller loop, there are two batteries and one resistor...
I don't know where to begin, any advice?
 

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