2 Ohm's law/Power equations questions

  • Thread starter Thread starter Normalization
  • Start date Start date
Click For Summary
SUMMARY

This discussion focuses on the application of Ohm's Law and power equations in analyzing electrical circuits, specifically addressing two questions related to current and brightness in lamps. The key equations referenced are V = IR, P = I²R, and P = V²/R. Participants confirm that the power output of a lamp is directly related to the current flowing through it, emphasizing that an increase in current results in increased brightness. The conversation also clarifies the behavior of current in parallel and series circuits, particularly when one lamp's filament breaks.

PREREQUISITES
  • Understanding of Ohm's Law (V = IR)
  • Familiarity with power equations (P = I²R, P = V²/R)
  • Knowledge of Kirchhoff's laws, particularly Kirchhoff's first law
  • Basic concepts of series and parallel circuits
NEXT STEPS
  • Study the implications of Kirchhoff's laws in circuit analysis
  • Explore the relationship between current and brightness in various electrical components
  • Investigate the effects of circuit modifications on total resistance and current distribution
  • Learn about practical applications of power equations in real-world electrical systems
USEFUL FOR

Students studying electrical engineering, educators teaching circuit theory, and anyone interested in understanding the principles of electricity and circuit behavior.

Normalization
Messages
23
Reaction score
0

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...
 
Physics news on Phys.org
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
 

Similar threads

Replies
16
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 7 ·
Replies
7
Views
6K