Alternating current

A coil rotating in a magnetic field will produce an e.m.f.

N

S

Rotating Coil Electrical Induction

This gif shows a conducting loop rotating through a magnetic field.

Flux through the loop is changing continuously, producing an alternating emf and current.

The slip rings at the ends of the loop allow it to be mechanically attached to a circuit using conducting brushes.

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Rotating Coil Electrical Induction

The gold arrows show the direction of velocity of the wires they are attached to.

When the arrows are perpendicular to the magnetic field, maximum change in flux occurs, hence: max emf.

You can think of the wires as “cutting” through the magnetic field at this time.

Cutting -> induced emf.

Rotating Coil Electrical Induction

When the arrows are parallel with the magnetic field, the change in flux is at a minimum.

Hence, for an instant, zero emf is induced.

There is no “cutting” of the magnetic field lines in this position, hence zero emf.

Rotating Coil Electrical Induction

If the arrow attached to the blue wire starts rotating clockwise from the initial position shown in the diagram, emf will be initially at a maximum.

At this starting position, current will be induced clockwise through the loop according to our RHR.

We then define that lockwise current produces a positive emf.

We can map it onto a position vs. vertical displacement graph:

Slip ring commutator

To use this e.m.f. to produce a current the coil must be connected to an external circuit using a slip-ring commutator.

Slip-rings

lamp

• In this position the magnetic flux is essentially zero but its rate of change is large. Hence, the voltage is at a maximum here and current flow is also at a maximum.

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• In this position the Voltage is now zero and their is no current flow. The magnetic flux is at a maximum but it hardly changes.

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Alternating current

The e.m.f. produced is sinusoidal (for constant rotation)

e.m.f.

V

Increasing the generator frequency?�Double frequency = double voltage

e.m.f.

V

Root mean square voltage and current

It is useful to define an “average” current and voltage when talking about an a.c. supply. Unfortunately the average voltage and current is zero!

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To help us we use the idea of root mean square voltage and current.

Root mean square voltage

e.m.f.

V

Root mean square voltage

First we square the voltage to get a quantity that is positive during a whole cycle.

e.m.f.

V

Root mean square voltage

Then we find the average of this positive quantity

e.m.f.

V

Root mean square voltage

We then find the square root of this quantity.

e.m.f.

V

Root mean square voltage

We then find the square root of this quantity.

e.m.f.

V

This value is called the root mean square voltage. The direct current equivalent

Root mean square voltage

We then find the square root of this quantity.

e.m.f.

V

V_rms = V_max/√2

E_max

Data booklet reference

• I_rms = I₀/√2
• V_rms = V₀/√2
• R = V₀/I₀ = V_rms/I_rms
• P_max = I₀V₀
• P = ½ I₀V₀

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11.2 AC questions

Transformers

What can you remember about transformers from IGCSE?

Transformers

How do they work?

N_p turns

N_s turns

V_p

V_s

Primary coil

Secondary coil

Iron core

“Laminated”

An alternating current in the primary coil produces a changing magnetic field in the iron core.

N_p turns

N_s turns

V_p

V_s

Primary coil

Secondary coil

Iron core

The changing magnetic field in the iron core induces a current in the secondary coil.

N_p turns

N_s turns

V_p

V_s

Primary coil

Secondary coil

Iron core

It can be shown using Faraday’s law that:

V_p/V_s = N_p/N_s and V_pI_p = V_sI_s (Power in = power out if ideal)

N_p turns

N_s turns

V_p

V_s

Primary coil

Secondary coil

Iron core

Data booklet reference

Ɛ_p/Ɛ_s = N_p/N_s = I_s/I_p

Power transmission

When current passes through a wire, the power dissipated (lost as heat) is equal to

​

P = VI across the wire

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Since V = IR

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Power dissipated = I²R

Power transmission

Power dissipated = I²R

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Since the loss of power depends on the square of the current, when transmitting energy over large distances it is important to keep the current as low as possible.

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However, to transmit large quantities of energy we therefore must have a very high voltage.

Power transmission

Electricity is thus transmitted at very high voltages using step up transformers and then step down transformers.

15,000 V

250,000 V

15,000 V

220 V

Transformer Substation

Residential

transformer

11.2 Transformer questions

Diode Bridges

Diode

• Allows (conventional) current to flow in one direction only (from + to – in the direction of the “arrow)

11.2 Investigating a diode bridge

This is a “compulsory” investigation

Half-wave rectifier

Half-wave rectifier with capacitor

Full-wave rectifier

When the input connected to the left corner of the diamond is positive, and the input connected to the right corner is negative, current flows from the upper supply terminal to the right along the red (positive) path to the output, and returns to the lower supply terminal via the blue (negative) path.

+

-

Full-wave rectifier

When the input connected to the left corner is negative, and the input connected to the right corner is positive, current flows from the lower supply terminal to the right along the red (positive) path to the output, and returns to the upper supply terminal via the blue (negative) path.

+

-

Diode bridge rectifier

Full-wave rectifier

Full-wave rectifier with capacitor

Diode bridge rectifier

Invented by Polish electrotechnician Karol Pollak and patent was recorded in 14 Jan, 1896