# Tapped Transformer With a Capacitor

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1. Jun 29, 2015

### Ahmed Bushra

The question is the following:

a) Determine all load voltages and currents
b) Determine the resistance reflected into the primary.

Answer a) R(l): 35V, 2.91A, X(cl): 15V, 1.5 A

The answer to part (b) is apparently 34.5 Ohms.

How did they arrive at these answers?

I am not sure whether both loads should be treated as parallel or in series. Since the polarity in the bottom half should be opposite, and hence the currents should be in the same direction (?).

How should the impedance/resistance in the secondary circuit be calculated?

I had ignored the impedance of the capacitor and arrived at a reflected resistance of 35.26 Ohms.

R(primary)=(1/n2) R(secondary)

Last edited by a moderator: Jun 29, 2015
2. Jun 30, 2015

### Staff: Mentor

You should be able to treat this as though the transformer has two independent secondary windings. This means the secondaries' loads are "in parallel".

3. Jun 30, 2015

### Ahmed Bushra

I see,

Well, in that case why would the voltage across one brach be different than the voltage across the other? (35 V and 10 V respectively)

Also, the resistance is:
12*10/10+12= 5.45 Ohms,

Primary Resistance = (1/n)2 ( Secondary Resistance)
n being the turns ration: 700/1200

Primary Resistance = 2.938 X 5.45 = 16.02

Why does the answer say 34.5 Ohms? Have I gone wrong somewhere?

4. Jun 30, 2015

### Staff: Mentor

What do you mean by "branch"?

What is the "10" in your resistance calculation? You need some brackets ( ) in your expression, too.

How many turns are you going to assign to each of the independent windings?

5. Jun 30, 2015

### rude man

You might want to approach this problem from a basic-physics viewpoint: all you need is Faraday's and Ampere's laws.
1. what is the flux (easy)?
2. so what are the output voltages? (easy too)
3. and the currents thru C and R? (easy too, bit of algebra required)
4. finally, and more challenging: what is the primary current? this will include a phase other than 0.

6. Jul 1, 2015

### Ahmed Bushra

By "branch" I mean each section of the parallel circuit. The first section being the one containing the resistor. The second section is containing the capacitor.

The "10" in the resistance calculation is the resistance of the capacitor. When I treat the secondary load as a resistor and a capactior in parallel: I have one resistor (12 Ohms) in parallel with a capacitor (10 Ohms), hence Resistance of the secondary load:

R= (R1*R2)/(R1+R2)
R=(10*12)/(10+12)= 5.45 Ohms

Why should I assign turns to the independent windings? I added them together, hence my ration of turns n=700/1200

Are you supposed to calculate each separately? If yes why? *confused*

7. Jul 1, 2015

### Ahmed Bushra

I can't workout the flux since they had not given any information with regards to area or magnetic field.

Last question is intriguing, what do you mean by "phase other than 0"?

8. Jul 1, 2015

### Staff: Mentor

If asked for the reflected resistance, you won't involve the capacitor---it contributes reactance but no resistance.

Model the transformer secondary as two independent windings, 700 turns powering the resistor, and 300 turns connected to the capacitor.

9. Jul 1, 2015

### The Electrician

Where did you get the answers? From the text?

You seem sure of the answers to part a), but for part b) you say "apparently". Are you uncertain of the answer for part b)?

10. Jul 2, 2015

### rude man

What's reflected into the primary is not a resistance but an impedance.

11. Jul 3, 2015

### rude man

You don't need area or B field. What does Faraday's law say?
I mean the primary current is not in phase with the primary voltage. The primary voltage phase is defined as zero.