# O Aim of the lecture  Inductors Behaviour in circuits Energy storage  Comparison with Capacitors V vs i Energy  Use in Circuits LR time constant sparks.

## Presentation on theme: "O Aim of the lecture  Inductors Behaviour in circuits Energy storage  Comparison with Capacitors V vs i Energy  Use in Circuits LR time constant sparks."— Presentation transcript:

o Aim of the lecture  Inductors Behaviour in circuits Energy storage  Comparison with Capacitors V vs i Energy  Use in Circuits LR time constant sparks  LCR Circuit Oscillations Filters o Main learning outcomes  familiarity with Inductors in circuits i and V LR curves Energy storage basic LC and LCR circuit Concept of filters Lecture 10

Reminder: o an inductor, as used in electronics is typically a coil of wire wound on a core The core increases the inductance by  r o Real inductors must have resistance because the wire has resistance o In fact they also have ‘stray’ capacitance, but we will ignore this inductor Resistor (resistance of wire in coil) L =  r  0 N 2 A/ l

Reminder:  For a resistor V=iR  For a capacitor i = C dV/dt For an inductor the relationship between current and voltage is: V = L di/dt Note the ‘similarity’ between the relationships for capacitors and inductors. As circuit elements they can be thought of as ‘opposites’ in the sense that i and V play opposite roles

For an inductor the relationship between current and voltage is: V = L di/dt So: o If the current is constant, then V=0  The inductor behaves simply like a (perfect) wire o If there is current flowing, then to reduce it to zero  requires a voltage as di/dt ≠ 0 [For a capacitor with a voltage on it, a current must flow for the voltage to reduce to zero, as Q=CV – similar]

So: o If the current is constant, then V=0  The inductor behaves simply like a (perfect) wire o If there is current flowing, then to reduce it to zero  requires a voltage as di/dt ≠ 0 Why? because there is energy stored in the inductor when it has a current flowing through it, Just like a capacitor has energy stored in it when it has a voltage across it. Work done = ∫ Power dt = ∫ iVdt = ∫ i(Ldi/dt)dt = ∫ iLdi = ½Li 2 Energy stored, E = ½Li 2 Capacitor Energy stored, E = ½CV 2

Energy stored, E = ½Li 2 Capacitor Energy stored, E = ½CV 2 Recall that there is NO energy stored in a resistor A resistor dissipates energy, it produces heat Inductors DO NOT dissipate any energy Capacitors DO NOT dissipate any energy These circuit elements can ABSORB energy But it can come out again, it is only stored

Where is the energy stored? In the magnetic field! There is an energy density associated with the field, so to create it work must be done. In a capacitor the energy is stored in the electric field, [but remember they are both aspects of the same e-m field]

 If the current flowing decreases, then  the stored energy must decrease,  the ‘excess’ must go somewhere.  It is absorbed by the circuit that is changing the current.  The inductor does work ON the circuit

This car relies on an inductor to make the (petrol) engine run - why? – it is used to make the spark to ignite the petrol

A petrol engine (not a diesel) needs a spark to ignite the petrol. An inductor is used. [in fact some modern cars do it differently, but still common] When the switch is closed, the voltage across and the current through the inductor will look like this

When the switch is closed, the voltage across, and the current through the inductor will look like this Because, Kirchoff’s law says V R + V L = E so iR + Ldi/dt = E this is a differential equation, the solution is i = ( E /R){1-e -t/(R/C) } time

o But if we now open the switch again,  there is a problem  no circuit path  so no current can flow o But the stored energy is = ½Li 2  so the current MUST flow  because there is energy stored o To keep current flowing  the inductor will generate a huge voltage,  big enough to break-down the air  cause a spark across the switch thousands of volts current will flow until energy dissipated it turns into heat and light

o But if we now open the switch again,  there is a problem  no circuit path  so no current can flow o But the stored energy is = ½Li 2  so the current MUST flow o To keep current flowing  the inductor will generate a huge voltage,  big enough to break-down the air  cause a spark across the switch thousands of volts current will flow until energy dissipated o Add a second gap, smaller than the switch gap It will break down first The smaller gap is the ‘spark gap’ Located where the spark is needed

Because:  The ‘spark gap’ in a car engine  is in the ‘spark plug’  screwed into the engine block  at top of cylinder gap Engineering Detail

Time after switch closed V L = V 0 e -t/(L/R) I L = (V 0 /R){1-e -t/(L/R) } V0V0 R L VLVL ILIL switch I or V The sparking situation is ‘abnormal’, it is more common to be using the components in a ‘controlled way. Sparks are usually bad!

Time after switch closed V c = V 0 e -t/CR I C = (V 0 /R){1-e -t/CR } V0V0 R C VCVC ICIC switch I or V Provided that L/R = RC then the shape of the I and V curves simply swap note ‘similarly’ in behaviour of L and C

Time after switch closed V L = V 0 e -t/(L/R) I L = (V 0 /R){1-e -t/(L/R) } V0V0 R L VLVL ILIL switch I or V Provided that L/R = RC then the shape of the I and V curves simply swap note ‘similarly’ in behaviour of L and C

Time after switch closed V c = V 0 e -t/CR I C = (V 0 /R){1-e -t/CR } V0V0 R C VCVC ICIC switch I or V Provided that L/R = RC then the shape of the I and V curves simply swap note ‘similarly’ in behaviour of L and C

Capacitors and Inductors are complementary devices (Often) o They can be ‘energised’ and ‘emptied of stored energy’ o This is called charging and discharging o (but recall this is NOT a statement about net electric charge) The relationship between voltage and current is: o V=iR resistors o V=Ldi/dt inductors o i=CdV/dt capacitors o Note that only resistors can ‘use’ energy  Dissipate it as heat Now consider using inductors and capacitors in together:

V0V0 L C VCVC I LC switch o In the circuit below, as shown,  the capacitor will energise (charge)  after a ‘long’ time V C = V 0.  I LC = 0 So the Energy stored in the circuit is E= ½CV C 2

V0V0 L C VCVC I LC switch o Now change the switch position, as shown below  the capacitor is energised, E = ½ CV C 2  V C = V 0.  I LC = L dI LC /dt ≠ 0 o But now there is an inductor connected across the capacitor o There is a voltage V C across L  Current must flow o As current grows, the voltage across capacitor will decrease  Because Q=CV and I = dQ/dt

V0V0 L C VCVC I LC switch o After some time, the capacitor will be fully de-energied  the capacitor has no energy, E = ½ CV C 2 = ½ C 0 2 = 0  V C = 0.  I LC = maximum = I LC max o As there is current flowing through an inductor o There is energy stored in its magnetic field  E L = ½ L(I LC max ) 2 o As no energy can be lost in this circuit (no resistance!)  E L = ½ L(I LC max ) 2 = ½ CV 0 2  so I LC max = V 0 √(C/L)

V0V0 L C VCVC I LC switch o All the energy is now stored in the inductor  E L = ½ CV 0 2  V C = 0 = V L  I LC =V 0 √(C/L) o As there is current flowing through the capacitor o It will energise (charge up)  The current will decrease  as the energy is transferred back to the capacitor o Eventually the current reaches zero and  E L = ½ L(I LC ) 2 = ½ L0 2 =0  E C = ½ CV 0 2

V0V0 L C VCVC I LC switch o The energy oscillates back and forth between  The inductor, stored as energy in magnetic field  The capacitor, stored as energy in the electric field  (and between the extremes, also shared between L and C) time I LC VCVC V 0 √(C/L) V0V0 The final answer is that the voltages and currents in this circuit oscillate like a sine wave But phase shifted

V0V0 L C VCVC I LC switch Write Kirchoff’s voltage law round the loop V = 0 = LdILC/dt + Q/C take derivative wrt to time V = 0 = Ld 2 I LC /dt + (1/C)dQ/dt = Ld 2 I LC /dt + I LC /C You will recognise this as SHM with  0 = 1/LC Or you should recognise it!!

V0V0 L C VCVC I LC switch 0 = Ld 2 I LC /dt + I LC /C I LC = {V 0 √(C/L)} sin(  0 t+  )  0 = 1/√(LC)  is found from initial boundary conditions: As I LC = minimum = 0 when t=0,  =  But  is not very important for this case, it just gives a definition of t=0  really matters when it is used to show the difference between the different voltages and currents in the circuit, so the shift between Current and voltage in capacitor is  /2 Similarly for the inductor, but it is –  /2 The voltages on the Capacitor and inductor are  different.

. In practice, all circuits have some resistance, so consider: The solutions are similar, with a sinusoidal oscillation, but now also with an exponential damping factor This is simply Damped SHM Identical equations to mechanics. (1) (4) Q = Q 0 e ‑ (t/  ) sin(  't +  ). I LC = dQ/dt  Close S1 to charge capacitor  Open S1 to disconnect battery  Close S2 to make circuit with L,R,C

VCVC VCVC Sinusoidal voltage source Filters: The combination of LC can be used to select frequencies (eg to choose a channel in a radio) All the analysis discussed has been ‘in the time domain’ This new kind of application, with AC signals (not batteries) is a whole new topic area, and called ‘analysis is in the frequency domain’. You don’t know everything yet! – but will need to take another course to find out all the clever things you can do in the frequency domain - engineering -

V out /V in frequency Some real filters Band-stop filter Enjoy!

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