Find transformer primary and secondary voltage/current

  • Thread starter Thread starter Bluestribute
  • Start date Start date
  • Tags Tags
    Transformer
Click For Summary

Discussion Overview

The discussion revolves around solving a transformer problem involving primary and secondary voltages and currents, with a focus on applying relevant equations and understanding the relationships dictated by the transformer’s turns ratio. The scope includes homework-related problem-solving and mathematical reasoning.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant initially assumes V1 to be 1000 Vrms but questions this after receiving unexpected results, suggesting the need to consider voltage division and the impact of series resistance.
  • Another participant points out that V1 cannot be 1000V due to voltage drop across the series resistance, referencing Ohm's Law.
  • There is a proposal to express the primary current in terms of other parameters to determine the voltage drop across the series resistance.
  • Some participants discuss the implications of the transformer being ideal with a 10:1 turns ratio, questioning how this affects the relationships between V1, V2, I1, and I2.
  • Participants suggest using Kirchhoff's Voltage Law (KVL) to derive equations that relate the variables involved, with some expressing uncertainty about the correct formulation of these equations.
  • There is a back-and-forth on whether the derived equations are correct, with one participant correcting another's sign in the equations.
  • One participant mentions finding an app to assist with calculations, indicating a struggle with the complexity of the problem despite its seemingly simple setup.

Areas of Agreement / Disagreement

Participants express differing views on the correct approach to solving for V1 and the relationships between the variables. There is no consensus on the correct equations or the initial assumptions about the voltages and currents.

Contextual Notes

Participants highlight the need for clear expressions for current and voltage to accurately apply the transformer equations, indicating potential limitations in their current understanding or assumptions about the problem setup.

Who May Find This Useful

Students working on transformer problems in electrical engineering or physics, particularly those dealing with voltage and current relationships in ideal transformers.

Bluestribute
Messages
192
Reaction score
0

Homework Statement


(in picture)

Homework Equations


V=IZ
All the ratio formulas (i.e. V2/V1 = N2/N1)

The Attempt at a Solution


I thought V1 would be 1000 Vrms. Well the answers are shown and it's not. So I thought maybe I would multiply Vrms (1000) by root 2 to find Vpeak (Vrms = Vpeak/root 2). Didn't work. I know if I find V1, the rest will fall into place because from there, it's simple ratios. But it's getting the answer of 790.6cos(18.43) that I'm having trouble with . . .
 

Attachments

  • ddd.jpg
    ddd.jpg
    57.9 KB · Views: 654
Physics news on Phys.org
V1 is not 1000V because some of Vs is lost in the 1000 ohm series resistance, by Ohms Law.
 
Right. But there's only one resistor, so the voltage always comes out to 1000. I tried doing voltage division this way:

V1 = (Vs(RL+XL))/(1010 + j20)

And didn't get the correct result either. I used the (10 + j20) combination as the load in my division.
 
You need to have an expression for the current in order to determine how much of Vs is lost across that series resistance. So what is an expression for the primary current in terms of other parameters?
 
As far as we can tell from the problem statement the transformer is ideal and has a 10:1 turns ratio. What does that tell us about the relationships between V1 and V2, I1 and I2?
 
NascentOxygen said:
You need to have an expression for the current in order to determine how much of Vs is lost across that series resistance. So what is an expression for the primary current in terms of other parameters?
Um . . . I don't know . . .

gneill said:
As far as we can tell from the problem statement the transformer is ideal and has a 10:1 turns ratio. What does that tell us about the relationships between V1 and V2, I1 and I2?

I know the ratio (V1/V2 = N1/N2, and inverse for current), but I need the V1 first . . .
 
Bluestribute said:
I know the ratio (V1/V2 = N1/N2, and inverse for current), but I need the V1 first . . .
That's why you need to write loop equations (KVL). The now known relationships between V1 and V2, I1 and I2, allows you to eliminate two of the variables, leaving two equations in two unknowns.
 
You can often start on the secondary side, and work back towards the primary.
 
NascentOxygen said:
You can often start on the secondary side, and work back towards the primary.
Go on . . . Because I thought of that but have no voltage or current to use.

gneill said:
That's why you need to write loop equations (KVL). The now known relationships between V1 and V2, I1 and I2, allows you to eliminate two of the variables, leaving two equations in two unknowns.
1000 + 1000I1 +V1 = 0
(10 + j20)I2 - V2 = 0

V1/V2 = I2/I1

These the equations you're referring to?
 
  • #10
Bluestribute said:
1000 + 1000I1 +V1 = 0
(10 + j20)I2 - V2 = 0

V1/V2 = I2/I1

These the equations you're referring to?
Almost. The first equation is not correct. Check the signs of the terms.

And you are given the transformer turns ratio, so use it.
 
  • #11
-1000 + 1000(I1) + V1 = 0
(10 + j20)I2 - V2 = 0

I2/I1 = 10 (Or N1/N2 if you'd prefer).

So I can turn I2 into 10I1 and V2 into 0.1V1?
 
  • #12
Bluestribute said:
-1000 + 1000(I1) + V1 = 0
(10 + j20)I2 - V2 = 0

I2/I1 = 10 (Or N1/N2 if you'd prefer).

So I can turn I2 into 10I1 and V2 into 0.1V1?
Yup.
 
  • #13
Boom. Took forever to find a Ti-84 App to do the trick but got one. Did not think you had to do that with such a simple transformer setup.
 
  • #14
Bluestribute said:
Boom. Took forever to find a Ti-84 App to do the trick but got one. Did not think you had to do that with such a simple transformer setup.
You need an app to solve an equation in one unknown?? :wink:
 
  • #15
Two unknowns. But when we only have 50 minutes for an exam, I need to find one that doesn't break my calculator! AHHH.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 12 ·
Replies
12
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 35 ·
2
Replies
35
Views
7K
  • · Replies 14 ·
Replies
14
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K