3. 8. 2003 1

IV–3 Energy of Magnetic Field

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Main Topics

• Transformers

• Energy of Magnetic Field

• Energy Density of Magnetic Field

• An RC Circuit

• An RL Circuit

• An RLC Circuit - Oscilations

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The Transformer I

• Transformer is a device with (usually) two or more coils sharing the same flux. The coil to which the input voltage is connected is called primary and the other(s) are secondary.

• Transformers are mostly used to convert voltages or to adjust (match) internal resistances.

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The Transformer II

• Let us illustrate the principle of functioning of a transformer on a simple type with two coils with N1 and N2 loops. We shall further suppose that there is negligible current in the secondary coil.

• Since both lops share the same magnetic flux, in each loop of each coil the same EMF 1 is induced:

1 = - dm1/dt

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The Transformer III

• If we connect a voltage V1 to the primary coil, the magnetization in the core will grow until the counter EMF induced in this coil is equal to the input voltage:

V1 = N11

• The voltage in the secondary coil is also proportional to its number of loops:

V2 = N21

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The Transformer IV

• So voltages in both coils are proportional to their number of loops:

V1/N1 = V2/N2

• It is more difficult to understand the case when the secondary coil is loaded and, of course, even much more difficult to design a good transformer with high efficiency. In big transformers it can be close to 1!

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The Transformer V

• Suppose, our transformer has efficiency close to 1.

• In this case, currents are inversely proportional to the number of loops in each coil and resistances are proportional to their squares:

P = V1I1 = V2N1I1/N2 = V2I2

I1N1 = I2N2

R1/N12 = R2/N2

2

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Energy of Magnetic Field I

• An inductance opposes changes in current. That means it is necessary to do work to reach a certain current in a coil. This work is transferred into the potential energy of magnetic field and the field starts to return it at the moment we want to decrease the current.

• If a current I flows through a coil and we want to increase it, we have to deliver power proportional to the rate of the change we want to reach.

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Energy of Magnetic Field II

• In other words we have to do work at a certain rate to move charges against the field of the induced EMF:

P = I = ILdI/dt dW = Pdt = LIdI

• To find the work to reach current I, we integrate:

W = LI2/2

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Energy Density of Magnetic Field I

• Similarly as in the case of a charged capacitor the energy here is distributed in the field, now of course magnetic field.

• If the field is uniform, as in a solenoid, it is easy to find the density of energy:

• We already know formulas for L and B:

L = 0N2A/l

B = 0NI/l I = Bl/0N

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Energy Density of Magnetic Field II

• Al is the inside volume of the solenoid, where we expect (most of) the field, can be attributed to the energy density of the magnetic field.

• This definition is valid in generally in every point of even non-uniform magnetic field.

AlB

N

Bl

l

ANU

0

22

0

20

2)(

0

2

2B

um

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RC, RL, LC and RLC Circuits

• Often not only static but also kinetic processes are important. So we have to find out how quantities depend on time when charging or discharging a capacitor or a coil.

• We shall se that circuits with LC will show a new effect – un-dumped or dumped oscillations.

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RC Circuits I

• Let’s have a capacitor C charged to a voltage Vc0 and at time t = 0 we start to discharge it by a resistor R.

• At any instant the capacitor can be considered as a power source and the Kirchhoff’s loop (or Ohm’s) law is valid:

I(t) = Vc(t)/R • This leads to differential equation.

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RC Circuits II

• All quantities Q, V and I decrease exponentially with the time-constant = RC

• Now, let’s connect the same resistor and capacitor serially to a power supply V0. At any instant we find from the second Kirchhoff’s law:

I(t)R + Vc(t) = V0

a little more difficult differential equation.

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RC Circuits III

• Now Q and V exponentially saturate while I exponentially decreases as in the previous case. The change of all quantities can be again described using the time-constant = RC.

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RL Circuits I

• A similar situation will be if we replace the capacitor in the previous circuit by a coil L.

• When the current grows the sign of the induced EMF on the coil will be the same as on the resistor and we can again use the second Kirchhof’s law:

RI(t) + LdI/dt = V0 • This is again a similar differential equation.

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RL Circuits II

• The coil refuses immediate growth of the current.

• I starts from zero and exponentially saturates. • The EMF on the coil (VL) starts from its

maximal value, equal to V0, and exponentially decreases. When the current becomes constant, the EMF on the coil disappears.

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LC Circuits I

• Qualitatively new situation appears when we connect a charged capacitor C to a inductance L.

• It can be expected that now the energy will change from the electric form to the magnetic form and back. We obtain un-dumped periodic movement.

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LC Circuits II

• This circuit is called an LC oscillator and it produces, so called, electromagnetic oscillations.

• We can use the second Kirchhoffs law:

L dI/dt – Vc = 0• This is again a differential equation but

of higher order.

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LC Circuits III

• What happens qualitatively:

• In the beginning the capacitor is charged and it tries to discharge through the coil. However EMF equal to the voltage on the capacitor builds on the coil to prevent quick growth of the current. The current is zero in the beginning. But its time derivative must be non-zero, so current slowly grows.

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LC Circuits IV

• The capacitor discharges which causes decreases the current growth and thereby also the EMF in the coil.

• At the point, the capacitor is discharged, the voltage on it and thereby the rate of the growth of the current as well as the voltage on the coil are zero. But the current is now in its maximum and the coil prevents it to drop instantly.

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LC Circuits V

• The EMF on the coil will now grow in the opposite direction to oppose the decrease of the current. But anyway the EMF as well as the decrease rate of the current grows. The capacitor will now be charged it the opposite polarity.

• At the moment the capacitor is fully charged the current is zero and everything repeats again.

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LC Circuits VI

• Using formulas for electric and magnetic energy we can find:

• Energy changes as expected from electric to magnetic and the total is constant.

)(cos22

22

02

tC

Q

C

QU e

)(sin22

22

022

t

QLLIUm

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LRC Circuits I

• If we add a resistor the oscillations will be dumped. The energy will be lost by thermal loses on the resistor.

• The level of dumping depends on the resistance.

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Homework

• No homework today!

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Things to read and learn

• This lecture covers:

• Chapter 26 – 4; 29 – 5, 6; 30 – 3, 4, 5, 6

• Advance reading:

• Chapter 25 – 6, 7; 29 – 4; 30 – 6; 31 – 1, 2

RC Circuit I• We use definition of the current I = –dQ/dt

and relation of the charge and voltage on a capacitor Vc = Q(t)/C:

RC

tQ

dt

dQ

R

tVtI c )()()(

• The minus sign reflects the fact that the capacitor is being discharged. This first order homogeneous differential equation can be solved by separating the variables.

RC Circuit II

• Where we define a time-constant = RC. We can integrate both sides of the equation:

dt

Q

dQ

• The integration constant can be found from the boundary conditions Q0 = CVc0 :

)exp()()ln( 0 t

QtQkt

Q

RC Circuit III

• By dividing this by C and then by R we get the time dependence of the voltage on the capacitor and the current in the circuit.:

)exp()( 0 t

CVtQ c

^

)exp()(

)exp()(

0

0

t

R

VtI

tVtV

c

cc

RC Circuit IV• We substitute for the current I = +dQ/dt

and the voltage and reorganize a little:

0

)(V

C

tQ

dt

dQR

• We get a similar equation for the charge on the capacitor but now its the right side is not zero. We can solve it by solving first a homogeneous equation and then adding one particular solution e.g. final Qk = CV0 .

RC Circuit V• Since we have already solved the

homogeneous equation in the previous case, we can write:

00 )exp()( CVt

QtQ

The integration constant we again get from the initial condition Q(0) = 0 Q0 = -CV0.

RC Circuit VI

• By dividing this by C we get the time dependence of the voltage on the capacitor:

)]exp(1[)( 0 t

CVtQ

)]exp(1[)( 0 t

VtVc

RC Circuit VII• To get the current we have to calculate from its

definition as the time derivative of the charge:

^

)exp()( 0

t

R

V

dt

dQtI

RL Circuit I

• First we solve a homogeneous equation (with zero in the right) and add a particular solution Im = V0/R (maximal current):

0)( Vdt

dILtRI

0)( dt

dILtRI

RL Circuit II• This can be solved by separation of the

variables. Using the previous, defining the time-constant = L/R and adding the particular solution, we get:

mIt

ItI

)exp()( 0 • We apply the starting conditions I(0) = 0

I0 = -Im and we get:

RL Circuit III

• The voltage on the coil we get from • V = LdI/dt :

)]exp(1[)(

tItI m

)exp()( 0 t

VtVL

^

LC Circuit I• We use definition of the current I = –dQ/dt

and relation of the charge and voltage on a capacitor Vc = Q(t)/C:

0)(

2

2

LC

tQ

dt

Qd

• We take into account that the capacitor is discharged by positive current. This is homogeneous differential equation of the second order. We can guess the solution.

LC Circuit II

• Now we get parameters by finding the second derivative of the Q(t) and substituting it into the equation:

)cos()( 0 tQtQ

• The solition are un-dumped harmonic oscillations.

LCtQ

LCtQ

10)(

1)(2

^