6.8 AC and transformers Flashcards Preview

A-Level Physics EDEXCEL Year 2 > 6.8 AC and transformers > Flashcards

Flashcards in 6.8 AC and transformers Deck (34)
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1
Q

What is alternating current?

A

Current that periodically changes direction

2
Q

What is direct current?

A

Current flowing only one way.

3
Q

What else can we have that is alternating?

A

voltage/pd

4
Q

What is an alternator? it uses what

A

generator of alternating current, uses electromagnetic induction

5
Q

how to alternators work?

A

as an electrical conductor is turned in a magnetic field an emf and this current is induced. Since the direction of rotation is switched the direction of emf/current in reversed. This is repeated

6
Q

What energy store is energy transferred to and from in alternators?

A

Ek to EPE

7
Q

What are transformers?

A

devices that use electromagnetic induction to change size of pd/voltage for alternating current supplies

8
Q

What does a transformer consist of?

A

An iron core, primary coil and secondary coil

9
Q

What are the 2 types of transformers?

A

Step Up and Step Down

10
Q

What is a step up transformer?

A

Where pd is increased due to the secondary coil having more coil turns than primary

11
Q

What is a step down transformer?

A

Where pd is increased due to the secondary coil having less coil turns than primary

12
Q

How does a transformer work?

A
  • The primary coil is supplied with a ac power supply and so creates its own magnetic field
  • the change is magnetic flux linkage relative to secondary coil means emf is induced in it (faradays law). Since AC is the supply this happens constantly as magnetic flux linkage keeps changing
  • The emf created is alternating since it would backtrack to oppose change of magnetic flux linkage. Its magnitude is determined upon the turns ratio.
13
Q

How do we determine the pd created in the secondary coil?

A

We use the ratios using the formula Vs/Vp = Ns/Np = Ip/Is
where v = voltage/pd/emf
n = turns
i = current

14
Q

Why is the current represented as Ip/Is in this equation instead of the other way round?

A

As it will do the opposite to voltage/pd (increase or decrease), since conservation of energy means power must stay the same. So IpVp = IsVs

15
Q

Why do transformers have an iron core? What is its structure?

A

A bunch of iron layers insulated from each other

It reduces the loss of energy transferred between coils

16
Q

Why are the layers of iron in the iron core insulated from each other?

A

Avoid heating effect and therefore energy loss as thermal energy

17
Q

Why would a DC not work for a transformer?

A

As teh magnetic flux linkage wouldn’t change so that no emf would be induced in the secondary coil (faradays law)

18
Q

What needs to decrease to increase voltage in secondary coil?

A

The current

19
Q

What needs to decrease to increase current in secondary coil?

A

voltage

20
Q

Why?

A

As power = IV must stay equal throughout to obey conservation of energy

21
Q

What is alternating current produced by ?

A

Alternating voltage

22
Q

What is the waveform (pd/t or i/t) for current and pd?

A

Sine graph

23
Q

What is the waveform for power of an ac supply?

A

Postive peaks only

24
Q

What happens to the the magnitude of power supply of an ac/av?

A

It won’t be the same as peak voltage/current or voltage/current of DC supply but lower as most of the time the pd/I will be below the peak.

25
Q

What do we need to do to compare ac/av to dc?

A

we have to use root mean square (r.m.s)

26
Q

What is root mean square?

A

voltage or current divided by (square root)2

27
Q

What can the root mean square be used for?

A

current

pd/voltage

28
Q

What is the r.m.s power of an ac supply?

A

Vr.m.s x Ir.m.s

29
Q

Using the waveform how can we work out the frequency?

A

Using the distance between peaks and f = 1/t

30
Q

How do we work out the r.m.s?

A

peak value divided by (square root)2

31
Q

How can we work out peak value from r.m.s?

A

r.m.s x (square root)2

32
Q

Is alternting current or emf constant?

A

No, varies from V0 to 0 each half a cycle

33
Q

How to work out power at a time?

A

p = V^2 / r = V0^2cos^2(omega x time)/R = V^2/2R

hence average power is half the peak power

34
Q

Equations for voltage and current for sine graphs?

A
V = V0 sin 2πft 
I = I0 sin 2πft 
f = frequency 
t = time  
2πf is essentially angular velocity

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