What is the correct equation to solve this power and voltage problem?

Click For Summary

Homework Help Overview

The discussion revolves around a power and voltage problem involving a toy motor connected to a transformer. The original poster is trying to determine the correct equation to find the primary current based on given values such as power, voltage, and current in the secondary circuit.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to use two different equations related to power and current in transformers but finds conflicting results. Some participants question the validity of the equations used and the assumptions made regarding the relationships between primary and secondary circuits.

Discussion Status

Participants are actively discussing the correct application of transformer equations and the implications of power conservation. There is a focus on clarifying the relationships between primary and secondary voltages and currents, with some guidance provided on the correct formulas to use. However, there is no explicit consensus on the resolution of the original poster's confusion.

Contextual Notes

The problem statement lacks clarity on what specifically is being solved, which has led to varied interpretations and approaches among participants. The original poster has indicated that they are unsure about the correct values and equations to apply, leading to further questioning of assumptions and definitions.

CAT 2
Messages
44
Reaction score
1

Homework Statement


A 4.0 Watt toy motor is connected to a secondary circuit of a transformer with 60 loops and a current of 0.10 A. The primary voltage is 120 V

Homework Equations



Vp / Vs = Ip / Is

Ip = Pp / Vp[/B]

The Attempt at a Solution



1. [/B]Vp / Vs = Ip / Is
120 V / 40 V = Ip / 0.10 A
0.3 A = Ip OR

Ip = Pp / Vp
Ip = 4 W / 120 V
Ip = 0.0333 AI have found people using both equations to solve this problem. Which one is correct? Did I just somehow put on the wrong values or something? I thought maybe the first one because in my course the second one was used to deal with circuits, and the first one was commonly used in the above type problems. But the above problem is sort of a circuit. Can anybody help me with this? It would be much appreciated!
 
Physics news on Phys.org
The statement of the problem does not specify what one is asked to solve for. Can you post that?
 
  • Like
Likes   Reactions: CAT 2
Sorry,

I am solving for current in the primary coil (Ip). I didn't know which equation was correct, because both could be used and they didn't come up with the same answer. I have already found that Vs = 40 volts.
 
Does that help?
 
CAT 2 said:
Does that help?
Your first solution is incorrect. The power (P = I V) is the same on both sides so you have Ip Vp = Is Vs.
This gives Ip×120 V = 40 V × 0.1 A → Ip = (4 /120) A = 0.0333 A, the same as the second solution.
 
So are you saying I messed up with my numbers? Or the power (P = I V) is the correct solution.
 
CAT 2 said:
So are you saying I messed up with my numbers?
I am saying you messed up your algebra, not your numbers. Your solution 1 says,
CAT 2 said:
1. Vp / Vs = Ip / Is
Where did you get this? If you cross-multiply, you get IsVp = IpVs which is incorrect. As I stated in post #5, the correct expression is Ip Vp = Is Vs.
 
This is what my course said, does that change anything?
ScreenShot_20180130161839.png
 

Attachments

  • ScreenShot_20180130161839.png
    ScreenShot_20180130161839.png
    36.2 KB · Views: 559
That is not the formula you used.in #1. Look carefully at the last line in your text and compare it with the first line of 3.1 in post #1.
 
  • Like
Likes   Reactions: CAT 2
  • #10
CAT 2 said:
This is what my course said, does that change anything?
It does not change anything. It just confirms that Ip Vp = Is Vs, what I already had in post #7.
 
  • Like
Likes   Reactions: CAT 2
  • #11
Thanks for all the help guys, I think I figured it out and 0.33333 is the proper answer. Am I right on this one?
 
  • #12
CAT 2 said:
Ip = 0.0333 A
CAT 2 said:
Thanks for all the help guys, I think I figured it out and 0.33333 is the proper answer. Am I right on this one?
Which one is right do you think?
 
  • #13
Woops typo, I meant 0.033333... Is that better?
 
  • #14
CAT 2 said:
Woops typo, I meant 0.033333... Is that better?
Yes, much better.
 
  • Like
Likes   Reactions: Merlin3189
  • #15
kuruman said:
Yes, much better.

I have the same question and no matter how I many times I try I always got 0.3 with the Ip/Is = Vp/Vs, and with the Ip/Vp = Is/Vs... I then did P=VI, which got 0.03 A, but the power I used in the equation was the power given for the primary circuit. Does that mean it won't work with secondary voltage and current, or is the power consistent throughout both circuits? I'm so confused!
 
  • #16
LonelyElectron said:
I have the same question and no matter how I many times I try I always got 0.3 with the Ip/Is = Vp/Vs, ...
No matter how many times you try with this equation you will get the wrong result because the equation itself is wrong.[/color] I have indicated as much in posts #5, #7 and #10.

The correct equation is Ip Vp = Is Vs which gives Ip /Is =Vs/Vp. You have Vp/Vs on the right side.
 
  • #17
kuruman said:
No matter how many times you try with this equation you will get the wrong result because the equation itself is wrong. I have indicated as much in posts #5, #7 and #10.

The correct equation is Ip Vp = Is Vs which gives Ip /Is =Vs/Vp. You have Vp/Vs on the right side.

Okay, so:
Ip/Is = Vp/Vs
Ip=VpIs/Vs
Ip=(120 V)(0.10 A)/40 V
Ip=0.3

Still getting 0.3... Am i manipulating the equation wrong?
 
  • #18
I'm afraid you are still getting it wrong. Compare your #17 with Kuruman's #16.

The equation Ip Vp = Is Vs is a statement of the primary power = secondary power. PP = PS

The equation with IP / IS and VP / VS is about the ratio of current and voltage in the primary and secondary.

But you keep saying they are the same. That is wrong. If the power stays the same, then if the voltage goes down, the current goes up and vice versa.
The ratio of currents is the inverse of the ratio of voltages.

So no t ## \frac {I_P} {I_S} = \frac {V_P} {V_S} \ \ but \frac {I_P} {I_S} = \frac {V_S} {V_P} ## The voltage ratio must be upside down (inverse) compared to the current ratio.

Does that help?
 
  • #19
LonelyElectron said:
Still getting 0.3... Am i manipulating the equation wrong?
No. You just didn't read my post #16 where I indicated that
LonelyElectron said:
Ip/Is = Vp/Vs
is wrong[/color][/size], yet you insist on using it. It does not follow from Ip Vp = Is Vs.
 

Similar threads

The efficiency of a transformer
  • · Replies 5 ·
Replies
5
Views
3K
Calculating turns in a primary coil of a transformer
  • · Replies 2 ·
Replies
2
Views
3K
Transformer question (between substation and home)
  • · Replies 1 ·
Replies
1
Views
1K
Brightness of lamp connected to transformer
  • · Replies 17 ·
Replies
17
Views
4K
Am I doing this ideal op-amp problem correctly?
  • · Replies 1 ·
Replies
1
Views
2K
Engineering Derive expressions for the voltage gain of this opamp circuit
  • · Replies 11 ·
Replies
11
Views
4K
What is the gain error for this opamp driven by a voltage
  • · Replies 18 ·
Replies
18
Views
3K
What voltage appears across the primary coil of this transormer
  • · Replies 3 ·
Replies
3
Views
6K
Very confusing question about transformer
  • · Replies 3 ·
Replies
3
Views
2K
Primary voltage and transformers
  • · Replies 2 ·
Replies
2
Views
4K