Year 13 Physics – Demonstrate understanding of electricity

8/4/2019 Year 13 Physics Demonstrate understanding of electricity

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External 6 Credits Achievement Standard 90523


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DC Electricity Define and use the concepts of current and

potential.

State and apply Ohms law.

Carry out calculations involving resistance,including resistances in series and inparallel.

Carry out calculations involving electrical

power. Say what internal resistance is, and carry out

related calculations.

Kirchoffs Laws State Kirchoffs current law, and explain how

it is a consequence of the conservation ofcharge.

State Kirchoffs voltage law, and explain howit is a consequence of the conservation ofenergy.

Carry out calculations involving the above.

Calculate the potential of one point in acircuit relative to another point.

Electric Fields and Capacitance. Define and electric field qualitatively.

Define the strength of an electric field.

State what is shown by the direction of anelectric field.

Give an example of a situation in whichhere is a uniform electric field.

Carry out calculations involving the workdone in moving a charged particle in auniform electrical field.

State and use 2 formulae for electrical fieldstrength, and state the 2 units for electricfield strength which corresponds to theseformulae.

Say what capacitance is and give thedefining equation.

Describe a parallel plate capacitor, say howcapacitance depends on three quantities,and state the relevant equation.

Carry out calculations involving storedenergy, capacitances in series andcapacitances in parallel.


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KCL (Kirchhoffs Current Law) For any point in a circuit, the current arriving

must equal the current leaving.

This is because charge can neither be created nor

destroyed; the amount of charge arriving at apoint in 1 s must equal the amount of chargeleaving from the point in 1 s.

+ = 3 +


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KCV (Kirchhoffs Voltage Law) The sum of the voltage changes around a closed

loop in a circuit is zero. This is because energy is conserved.

Any charge which leaves a point with a certainamount of electrical energy must on its return tothat point have the same amount of electrical

energy. Remember =

+ + + = 0

8 + 10 7 6 3 14 = 0


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Generateequations usingKCL and KVL.

Solve them to

find the unknownquantities.

You need thesame number of

distinctequations as youdo unknowns.

KVL tricks: Pick a starting point and go

around the loop. The voltage across a

resistor is positive if you aregoing around the loop inthe same direction as thecurrent through the resistor

The voltage of a battery is

positive if you come to itspositive plate before itsnegative plate.


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Every voltage can be

considered to be acombination of aelectromotive force, ,and an internalresistance, .

If we measured the voltage of the circuit when nocurrent is flowing, then we would measure theelectromotive force, EMF.

If we measured the voltage of the circuit when thereis a current is flowing, then we would measure the

voltage that is reduced by the internal resistance. This yields the equation:

=


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Mutual inductance occurs when the changing

magnetic flux induces a voltage in a nearbycoil, and thus causes a current in that coil.

Self inductance is when this occurs in thesame circuit, and even in the same coil! When a current flows through a circuit, it produces

a magnetic field.

This produces magnetic flux through the circuit.

If the current changes, the flux through the circuit

changesthis change of flux produces a voltage inthe circuit.


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The flux in the circuit will be

proportional to the currentthat causes it: =

(L is the constant ofproportionality)The voltage is given by:

=

Combining these equations:

=

L (unit, henries, H) is calledthe self inductance of thecircuit, often shortened to

just inductance.

The equation shows that the

self inductance acts verystrangely:

If the current is increasing: The induced voltage will

oppose the increase.

If the current is decreasing: The induced voltage will try

to keep the current going.

If the currrent is constant: The induced voltage is zero

there is no voltage across theinductance.


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In this circuit, if we were tomeasure the voltages across Cand R, we would find:

+

This equation would not hold forRMS or maximum voltage values.

When would it hold true?

We can say = + for anyinstantaneous measurement of thevoltages. So how can we deal with this

problem?

R

C

A


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We start by drawing a

reference circle:

We know that the alternating current andthe voltage through the resistor will be inphase (they occur at the same time).

I

VR

We also know that the capacitors voltage

is

radians behind the alternating current.

VC

The alternating voltage is therefore foundby adding + . This will give us .

VA

This phasor diagram will give this graphwhen rotated. Each instant of the graph isaccurate.


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The next step here is

to realise how muchwe can calculate fromthis vector diagram. VR

VC

VA

Because it is a rightangled triangle, weuse Pythagoras to

calculate the size of.

= +

= +

= +

1

Here we have used = forthe resistor and

=

for

the capacitor. Remember =

. Now lets factorise.

This equation looks a lot like = , butthe resistance is now a combination ofthe reactance and the resistance.


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When we use a vector diagram like this to combinethe reactance and the resistance, we get what is calledthe impedance of a circuit.

Impedance, Z, is measured in ? Ohms! it is still basically a resistance. For a capacitor and a resistor in series in an

alternating current circuit, the impedance is given by

= +

Now we can write = Just to clarify: Reactance is the effective resistance of

a capacitor or inductor in an alternating currentcircuit. Impedance is the vector combination of the resistance

and the reactance, taking into account the phase ofeach reactance.


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We also need toknow how muchphase differencethere is on .

The phasedifference angle ismeasured between

and

. Why?

Simple trigonometry:

= tan

VR

VC

VA

= tan

= tan1


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We can perform a verysimilar derivation withan inductor in serieswith a resistor.

The main difference isthat the inductancevoltage phasor leadsthe voltage through

the resistor by

.

R

L

A

Recall the reactance ofan inductor is givenby = .


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The impedance of a RLcircuit running analternating current:

= +

= +

= +

= +

So now we can write =

VA

VL

VR

= tan

= tan


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Capacitor Inductor

The impedance of anRC circuit:

= +1

The phase difference:

= tan1

The impedance of anRL circuit:

= + The phase difference:

= tan


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120

0.68F

A

The RC circuit opposite shows a 120 resistor in series with a 0.68 Fcapacitor. The supply voltage is set

to be 6.0 VRMS, at a frequency of 2.4kHz.

1. Calculate the angular frequency ofthe supply.

2. Calculate the reactance of thecapacitor.

3. Write down the reactance of theresistor.

4. Sketch a vector diagram of the

reactances, and use this to calculatethe impedance of the circuit.5. Use Ohms Law to calculate the RMS

current drawn from the supply.


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120

540 mH

A

The RL circuit opposite shows a 120 resistor in series with a 540 mHinductor. The resistor voltage is

measured to be 1.5 VRMS and theinductor voltage is measured to be2.0 VRMS.

1. Sketch a voltage phasor diagram.2. Sketch a voltage vector diagram,

and use it to calculate the voltageof the AC supply.

3. Calculate the reactance of theinductor.

4. Calculate the impedance of the

circuit.5. Calculate the generatorfrequency.


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Now we are going tocombine a resistor, aninductor and a capacitor inseries in the same circuitwith an alternating current

power supply. Yes, its kind of epic.

Recall:

is in phase with thecurrent.

is

behind the current.

is

ahead of the

current.


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VR, VCand VL aredrawn as shown.

VA

VL

VR

VC

You can see that VCand VL are directly

opposite eachother, and willcancel vectorially.

VL-VC

The A.C. voltage is

the vectorialcombination ofVRand VL VC.


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So we can see that:

= +

= +

= +

= 1

+

= 1

+ This is the impedance of

the LCR circuit!

Using the reactancesof L and C.

Factorise the currentout.


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This is the impedanceof the LCR circuit.

We can see that theimpedance isminimised when

= 0, or =

= 1

+

In fact, when =

the impedance of the

circuit simplifies to =

, or = . This means that under certain conditions,

an AC circuit can be considered to haveonly resistance.


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There are 3 ways make an AC circuit resonate. We need to make =

1. We could change L.2. We could change C.

3. We could alter the frequency. Given that = 2, and that the resonant

frequency occurs when =

, solve for

the frequency.

That last one was an instruction, I actuallywant you to try to find the resonantfrequency equation.


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= 1

=1

=

1

=1

2 =1

=1

2

This frequency is called theresonant frequency. At resonance, the effects of the

inductance and the capacitancecancel out and the circuit

impedance equals the resistance. When the frequency is lower than

the resonant frequency, thecapacitance effects dominate thecircuit.

When the frequency is higher thanthe resonant frequency, theinductance effects dominate thecircuit.


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Note that this is called the

resonant frequency for areason. It can resonate.

For example, in this circuit:

= 1

= 40 mH = 25 F = 20 V

Calculate the resonant

frequency. Calculate , , and at

this frequency.

At resonance, theresistor is the onlything limiting thecurrent.

This means the

voltages across theother componentscan be much largerthan the supplyvoltage.

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Year 13 Physics – Demonstrate understanding of electricity