2 Ohm's law/Power equations questions

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Homework Help Overview

The discussion revolves around two questions related to Ohm's law and power equations, specifically focusing on the relationships between voltage, current, and power in electrical circuits.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants explore the reasoning behind selecting specific answers based on power equations and their implications on circuit behavior. There are discussions on the relationship between current and brightness in lamps, and whether it is necessary to use power equations or if Ohm's law suffices. Some participants question the consistency of current through different components in the circuit.

Discussion Status

Participants are actively engaging with each other's reasoning, providing clarifications and corrections. There is recognition of the need to describe current behavior more carefully in relation to circuit changes. Some guidance has been offered regarding the interpretation of current flow before and after changes in the circuit.

Contextual Notes

There are mentions of specific circuit configurations, such as parallel and series arrangements, and how these affect current distribution. Participants also refer to a typo regarding the equations used, indicating a focus on accuracy in mathematical representation.

Normalization
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Homework Statement


I've done both these questions with correct answers, but I just want sure that I chose the right choices for the right reasons. The questions are:
PhysQ.png



Homework Equations



V = IR
P = I2R
P = V2/R

The Attempt at a Solution



The answers for both are D.

The reason I chose D for the first one is because P/I2 = R And due to Ohm's law R is constant, so I2 = P So the graph should look like a y = x2 graph.

The reason I chose D for the second one is because the brightness of the lamp depends on the power output so since P = I2R an increase of I (Due to Kirchovs first law) across L would mean a larger power output across L, and a decrease in I (because of the increase in R across the parallel part) across N would mean due to the same equation that P would decrease. For this particular question is it 100% necessarily to use the power equations or could you just say that a larger current means a greater brightness and then just deduce the answer from Ohm's law because the Examiner's report didn't mention anything about using the power equations...
 
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Hi Normalization! :smile:
Normalization said:
The reason I chose D for the first one is because P/I2 = R And due to Ohm's law R is constant, so I2 = P So the graph should look like a y = x2 graph.

But there's no I on those graphs, only V … you need an explanation that uses V instead. :redface:
The reason I chose D for the second one is because the brightness of the lamp depends on the power output so since P = I2R an increase of I (Due to Kirchovs first law) across L would mean a larger power output across L, and a decrease in I (because of the increase in R across the parallel part) across N would mean due to the same equation that P would decrease. For this particular question is it 100% necessarily to use the power equations or could you just say that a larger current means a greater brightness and then just deduce the answer from Ohm's law because the Examiner's report didn't mention anything about using the power equations...

Technically, you're correct … brightness means power …

but in practice we just use the fact that (for a particular component) that brightness increases with current, so no you should just use current.

The current is not the same through each lamp, so you'll have to describe what happens to the currents more carefully.
 
tiny-tim said:
The current is not the same through each lamp, so you'll have to describe what happens to the currents more carefully.

It should be clarified that the current going through each lamp is not the same before the filament of lamp M breaks. Once the filament of lamp M breaks, you are no longer dealing with a circuit containing parallel resistors (where the current varies, but the voltage remains constant), but with a pair of resistors in a series (where the current remains constant, but the voltage varies).

Hope this helps :smile:
 
Thanks a lot guys that really helps!

I meant V^2/P sorry not P/I^2 (typo). And yes the current stays constant throughout the whole circuit (after removal of parallel light bulb) and increases relative to L and decreases relative to N.
 
Hi Normalization! :smile:
Normalization said:
And yes the current stays constant throughout the whole circuit (after removal of parallel light bulb) and increases relative to L and decreases relative to N.

Why?

Describe what the currents are before and after (with numbers).
 
Before:

RTotal = 3R/2 Because of R + 1/(1/R + 1/R) = R + R/2 = 3R/2
So ITotal = V/(3R/2) = 2V/3R
IL = ITotal/2 (identical lamps) = 2V/6R = V/3R
IN = ITotal = 2V/3R

After:

RTotal = 2R Because RTotal = R1+R2 = R+R = 2R
So ITotal = V/2R
IL = IN = ITotal = V/2R
V/2R < 2V/3R So IN Decreases
V/2R > V/3R So IL Increases
 
excellent! :smile:

that's the way to do it! :wink:
 
yee :D
 
BTW: If you have 2 resistors in parallel (R1 and R2), do you just find the current going through one resistor R1 by doing R1/(R1 + R2) × ITotal?
 
  • #10
yes :smile:

(can you prove that? :wink:)​
 
  • #11
Are you sure it's not I1 = R2 / (R1 + R2) IT? Because:

I1 = VP/R1 Where VP is the potential difference across the parallel part. And VP = IT × 1/(1/R1+1/R2) And 1/(1/R1+1/R2) = 1/(R1+R2/R1×R2) = (R1×R2/R1+R2).

So VP = IT × (R1×R2/R1+R2)

So I1 = IT × (R1×R2/R1+R2) × 1/R1 = IT × (R1×R2/R1(R1+R2)) = IT × R2/(R1+R2)
 
  • #12
oh, I didn't see the R1 at the end of the line in your previous post …

it was off the end of my screen :redface:

yes, that was the correct formula for I2

your new formula is the correct formula for I1
 
  • #13
Awesome, thanks :P
 

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