# Transformer - Calculating Output Voltage (AC Input Voltage)

## Homework Statement

For a transformer, the ratio number of turns on the transformer secondary : number of turns on the transformer primary = 21 : 1200. If the input voltage is 120V AC, what is the output voltage?

## Homework Equations

$\frac {N_{P}}{N_{S}}=\frac {V_{P}}{V_{S}}$

## The Attempt at a Solution

I used the equation above to calculate the output voltage:
$\frac {1200}{21}=\frac {120}{x}$. So the output voltage, which is x, is 2.1V
However, the answer of the book is 4.2V. I wonder if I have to double the value of the output voltage to get the final answer (the peak-to-peak output voltage). But I think the transformers always work with AC current and the formula above has already been constructed with AC current. Can anyone please correct me if I am wrong?

I appreciate any help given! Thank you!

NascentOxygen
Staff Emeritus
I used the equation above to calculate the output voltage:
. So the output voltage, which is x, is 2.1V
However, the answer of the book is 4.2V.
Your method is correct, so the book is wrong.

Unless stated otherwise, AC voltage and current is always taken to be the RMS value.

Thank you NascentOxygen. You cleared my doubt!

Can you please explain why we take RMS values when dealing with transformers? In my textbook, the sample problems (about transformers) just simply state the voltage values (e.g. A step-down transformer is designed to convert 240 V AC to 12 V AC).

Thank you very much again!

tiny-tim
Homework Helper
Hi LovePhys! Can you please explain why we take RMS values when dealing with transformers?

I think it's because it makes average power easy to calculate …

Pav = VrmsIrms (or Pav = VrmsIrmscosφ if V and I are out of phase by φ)

NascentOxygen
Staff Emeritus
Thank you NascentOxygen. You cleared my doubt!

Can you please explain why we take RMS values when dealing with transformers?
I didn't mean to imply when dealing with transformers, I meant all the time. RMS is the convention for alternating voltages and currents for energy purposes. Probably originated because, e.g., 10V RMS (of any waveshape) gives the same heating effect as 10VDC. It is easy to relate AC RMS with the equivalent in DC. You can run a 120V AC incandescent light bulb off 120VDC.

X volts DC has an RMS value of X VRMS.

Thank you very much indeed, NascentOxygen and tiny-tim! I learnt a lot from you two!

LovePhys

rude man
Homework Helper
Gold Member

## Homework Statement

For a transformer, the ratio number of turns on the transformer secondary : number of turns on the transformer primary = 21 : 1200. If the input voltage is 120V AC, what is the output voltage?

## Homework Equations

$\frac {N_{P}}{N_{S}}=\frac {V_{P}}{V_{S}}$

## The Attempt at a Solution

I used the equation above to calculate the output voltage:
$\frac {1200}{21}=\frac {120}{x}$. So the output voltage, which is x, is 2.1V
However, the answer of the book is 4.2V. I wonder if I have to double the value of the output voltage to get the final answer (the peak-to-peak output voltage). But I think the transformers always work with AC current and the formula above has already been constructed with AC current. Can anyone please correct me if I am wrong?

I appreciate any help given! Thank you!

Sounds like your transformer has a center-tap (CT). If in the primary, the 120V is applied from one end to the CT. If in the secondary, then the turns ratio you were given is that of primary to one-half the secondary.

Hello rude man,

In my textbook, they mentioned transformer and the formula I used in this problem. Actually, I also studied about center-tap full-wave rectifier, but the book just explains what happens when the secondary coil provides an input voltage to the circuit.

Sounds like your transformer has a center-tap (CT). If in the primary, the 120V is applied from one end to the CT. If in the secondary, then the turns ratio you were given is that of primary to one-half the secondary.
I'm not quite sure what you mean. But from my understanding, I think that the transformer in a center-tap rectifier basically will work in the same way as a normal transformer.

I am a high school student, so please forgive me if I'm wrong!
Thank you!

Sorry, I'm unable to edit my previous post.

I think I know what you're talking about, rude man. The center tap will split the secondary part in two equal halves (both the number of turns and the voltage). I agree that the problem didn't state clearly which kind of transformers is, but from sample problems in the book, I think this is just a normal transformer. Thank you, rude man. I learnt a new thing!