# Voltage in secondary coil of transformer

1. May 25, 2016

### Physgeek64

1. The problem statement, all variables and given/known data
A Primary coil, with $n_1$ turns, is driven with a voltage $V=V_0sin(wt)$

Find the voltage in a secondary coil, with $n_2$ turns, as a function of time

2. Relevant equations

3. The attempt at a solution
Mutual Inductance $M=\mu_0 n_1 n_2/l *area$
$emf=-d\phi/dt$
$\phi=\mu_0 n_1 n_2 area/l *I$

$v=LdI/dt$
$I=1/L \int v dt$
$I=1/L (-1/w cos(wt))$

I don't know whether I'm allowed to use the usual expression here for relating current and voltage in an inductor, since we have mutual inductance?

I don't really know how to find the current in the primary coil, and hence can't find the voltage across the secondary coil.

My other thought was: am I allowed to use the transformer equation, or does this only apply for max / rms voltages?

Many thanks

Last edited by a moderator: May 26, 2016
2. May 26, 2016

### andrevdh

Isn't it just the turns ratio that governs the induced voltage in the secondary?

3. May 26, 2016

### Physgeek64

I get that this gives the peak voltage ratio. But does this hold in terms of their phases? I.e. if one is driven with no initial phase, does the current in the secondary coil also have no initial phase? :) Thanks

4. May 26, 2016

### andrevdh

According to mutual induction theory the induced voltage in the secondary coil is directly proportional to the negative of the rate of change of the current in the primary coil.

5. May 26, 2016

### Physgeek64

Ah of course. So this would lead to a phase change of $\pi/2$ ?

6. May 26, 2016

### Merlin3189

If you are using typical iron cored power transformer with small leakage reactance, the same flux links both primary and secondary, so the induced voltage per turn in the primary is the same as the induced voltage per turn in the secondary. No phase change.

7. May 27, 2016

### rude man

What does "no phase" mean? Let's say "zero phase" instead.

The phase of the secondary winding depends on what you label the "low" end and the "high" end. If winding "dots" are facing each other there is no phase change, otherwise it's 180 degrees.

You'll have to ask someone or look it up to learn what "dotted ends" means, probably.
Also, there is no secondary winding current without a secondary load.

8. Jun 5, 2016

### Physgeek64

So I looked up the dot convention, and I understand it as a rule of thumb, but why does this happen. Suppose you wind them such that their dotted ends are either both up or both down (i.e. there will be no phase lag between the secondary and primary voltages), if you work through the maths, you still only take one derivative to calculate the voltage across the secondary coil given the voltage in the primary. Supposing the primary voltage is sinusoidal, then the secondary voltage should be co-sinusoidal.... which is a phase change of $\frac{\pi}{2}$ ?

Many thanks :)

9. Jun 5, 2016

### rude man

The primary voltage is 90 degrees out of phase with the primary flux: emfp = dΦ/dt. Then, the voltage induced in the secondary winding is also 90 degrees out of phase with the same flux: emfs = dΦ/dt. So 90 + 90 = 0 or 180, depending on how the dotted ends face.

10. Jun 5, 2016

### Physgeek64

Oh okay, so theres also a phase shift in the primary circuit! Thank you, that helps a lot :)

11. Jun 5, 2016

### rude man

Right! There's a phase shift in the primary circuit because the magnetic flux phase is 90 degrees off the primary voltage.
BTW this assumes an ideal transformer with no winding resistance and leakage flux.

12. Jun 5, 2016

### Physgeek64

Why is it that the flux is out of phase- is this simply by virtue of the inductor requiring time to build up the B-field? (Excuse my sloppy wording)

13. Jun 5, 2016

### rude man

The flux is in-phase with the input current, but the input current lags the input voltage by 90 degrees.

The definition of inductance is V = L di/dt! And you know that d/dt sin(wt) = w cos(wt), right?

14. Jun 5, 2016

### Physgeek64

Oh okay! Thank you