2 Ohm's law/Power equations questions

  • Thread starter Normalization
  • Start date
In summary: I2In summary, the first equation states that the total current is equal to the sum of the currents going through the two resistors in parallel. The second equation states that the current through lamp M decreases after the filament of lamp M breaks.
  • #1
Normalization
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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...
 
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  • #2
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.
 
  • #3
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:
 
  • #4
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.
 
  • #5
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).
 
  • #6
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
 
  • #7
excellent! :smile:

that's the way to do it! :wink:
 
  • #8
yee :D
 
  • #9
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
 

Related to 2 Ohm's law/Power equations questions

What is Ohm's Law?

Ohm's Law is a fundamental law in physics that describes the relationship between voltage, current, and resistance in an electrical circuit. It states that the current through a conductor is directly proportional to the voltage and inversely proportional to the resistance.

What is the equation for Ohm's Law?

The equation for Ohm's Law is: I = V/R, where I is the current in amperes (A), V is the voltage in volts (V), and R is the resistance in ohms (Ω).

What is the power equation?

The power equation is: P = VI, where P is the power in watts (W), V is the voltage in volts (V), and I is the current in amperes (A). This equation is used to calculate the amount of electrical energy consumed or produced in a circuit.

How do I calculate resistance using Ohm's Law?

To calculate resistance using Ohm's Law, R = V/I. Simply divide the voltage by the current to get the resistance in ohms (Ω).

What is the difference between series and parallel circuits?

In a series circuit, the components are connected in a single loop, with the same current passing through each component. In a parallel circuit, the components are connected in branches, with each branch having its own current. This results in different calculations for voltage and current in each type of circuit.

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