---

With good reason.

Since each of the supplies is connected to the transformer as a

full-wave center-tapped supply, the rectifiers will only drop about

0.7V in front of the filters, so with 27VRMS into the rectifiers the

peak voltage (the voltage across each filter cap) will be:

Vp = (Vrms * sqrt(2)) - 0.7V = 37.7V

Which is pretty close to what you got.

---

---

Since a transformer transfers power and the secondary can deliver:

P = IE = 1.25A * 24V = 30 watts

into a resistive load, then the primary will need to take 30 watts

from the mains in order to transfer that into the load.

Then, since:

P= IE

we can rearrange to solve for the primary current:

P 30W

I = --- = ------ = 0.25A

E 120V

for 120VRMS mains.

Assuming your transformer is 80% efficient means that the mains will

have to supply extra current to compensate for the transformer's

losses, so that means you'll need to draw about

0.25A

It = ------- ~ 0.31 amperes

0.8

from 120V mains.

---

I took a new measurement and found out the the ac value was acutally 27V not

24. So when i muliply with the square of two i roughly get 37,5V peak to

peak.

But why do i see the peak to peak voltage after it has been rectified???

Even with load 40ohms load.

---

Unless you're measuring the difference in voltage between the filter

caps you're not measuring peak-to-peak, you're measuring peak.

Since each of the supplies is connected to the transformer as a

full-wave center-tapped supply, the rectifiers will only drop about

0.7V in front of the filters, so with 27VRMS into the rectifiers the

peak voltage (the voltage across each filter cap) will be:

Vp = (Vrms * sqrt(2)) - 0.7V = 37.7V

Which is pretty close to what you got.

You don't say whether the load is across the filter cap or across the

output of the regulator but, assuming it's across the filter cap, the

current through the resistor will be

E 37.7V

I = --- = ------- ~ 0.94 amperes

R 40R

This will cause the voltage across the filter to vary as it charges up

through the rectifier and discharge when the voltage out of the

rectifier falls below the voltage on the cap. This voltage change is

called 'ripple', and its amplitude varies with the load current, like

this:

I dT

dV = ------

C

where dV is the ripple voltage in volts,

I is the load current in amperes,

dt is the ripple period in seconds, and

C is the capacitance in farads

So,

0.94A * 8.3E-3s

dV = ---------------- = 1.66 volts

4.7E-3F

That means that that the caps could charge up to 37.7V, but that that

voltage would fall to about 36V when the rectifier output fell to

lower than that. Your DC voltmeter would probably read the average

of those two voltages, about 36.8V, so you probably wouldn't even

notice the difference between loaded and unloaded unless you were

intent on noticing it.

If the 40 ohm load was connected across the output of a regulator, the

current would be:

15V

I = ----- = 0.375A

40R

and the ripple voltage would fall to:

0.375A * 8.3E-3s

dV = ------------------ ~ 0.66 volts

4.7E-3F

so you'd probably be even less likely to notice that difference.

There are a few more problems to consider, but in order to do that we

need to know what your maximum load current is going to look like.