# AC Circuits and Resonance Conclusion

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5 Οκτ 2013 (πριν από 4 χρόνια και 8 μήνες)

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AC Circuits and Resonance

Conclusion

April 20, 2005

Alternating Current Circuits

w

is the
angular frequency

w

we use the
frequency

f
[cycles per second]

Frequency

f
[cycles per second, or Hertz (Hz)]
w

= 2p

f

V = V
P

sin (
w
t
-

f
v
)

I = I
P

sin (
w
t
-

f
I
)

An “AC” circuit is one in which the driving voltage and

hence the current are sinusoidal in time.

p

f
v

2p

V(t)

w
t

V
p

-
V
p

p

f
v

2p

V(t)

w
t

V
p

-
V
p

V = VP sin (wt
-

f
v

)

Phase Term

Resistors in AC Circuits

E

R

~

EMF (and also voltage across resistor):

V = V
P

sin (
w
t)

Hence by Ohm’s law, I=V/R:

I = (V
P

/R) sin(
w
t) = I
P

sin(
w
t)

(with I
P
=V
P
/R)

V and I

“In
-
phase”

V

w
t

p

I

2p

This looks like I
P
=V
P
/R for a resistor

(except for the phase change).

So we call
X
c

= 1/(
w
C)

the

Capacitive Reactance

Capacitors in AC Circuits

E

~

C

Start from:
q = C V
[V=V
p
sin(
w
t)]

Take derivative: dq/dt = C dV/dt

So I = C dV/dt = C V
P

w

cos (
w
t)

I = C
w

V
P

sin (
w
t +
p
/2)

The reactance is sort of like resistance in
that
I
P
=V
P
/X
c
the voltage by 90
o

(phase difference).

V

w
t

p

2p

I

V and I “out of phase” by 90
90
º.

What the **(&@ does
that

mean??

I

V

Current reaches it’s maximum at

an earlier time than the voltage!

1

2

I = C

w

V
P

sin (
w

p
/㈩
=
f

Again this looks like I
P
=V
P
/R for a

resistor (except for the phase change).

So we call
X
L

=
w
L

the

Inductive Reactance

Inductors in AC Circuits

L

V = V
P

sin (
w
t)

Loop law: V +V
L
= 0 where V
L
=
-
L dI/dt

Hence: dI/dt = (V
P
/L) sin(
w
t).

Integrate: I =
-

(V
P
/ L
w)

cos (
w
t)

or

I = [V
P

/(
w
L)] sin (
w
t
-

p
/2)

~

Here the current lags the voltage by 90
o
.

V

w
t

p

2p

I

V and I “out of phase” by 90
º. I lags V by
90
º.

Phasor Diagrams

V
p

I
p

w
t

Resistor

A phasor is an arrow whose length represents the amplitude of

an AC voltage or current.

The phasor rotates counterclockwise about the origin with the

angular frequency of the AC quantity.

Phasor diagrams are useful in solving complex AC circuits.

The “y component” is the actual voltage or current.

Phasor Diagrams

V
p

I
p

w
t

V
p

I
p

w
t

Resistor

Capacitor

A phasor is an arrow whose length represents the amplitude of

an AC voltage or current.

The phasor rotates counterclockwise about the origin with the

angular frequency of the AC quantity.

Phasor diagrams are useful in solving complex AC circuits.

The “y component” is the actual voltage or current.

Phasor Diagrams

V
p

I
p

w
t

V
p

I
p

w
t

V
p

I
p

w
t

Resistor

Capacitor

Inductor

A phasor is an arrow whose length represents the amplitude of

an AC voltage or current.

The phasor rotates counterclockwise about the origin with the

angular frequency of the AC quantity.

Phasor diagrams are useful in solving complex AC circuits.

The “y component” is the actual voltage or current.

+

+

i

+

+

+

+

i

i

i

i

i

LC Circuit

time

Analyzing the L
-
C Circuit

Total energy in the circuit:

Differentiate :

N o change
in energy

Analyzing the L
-
C Circuit

Total energy in the circuit:

Differentiate :

N o change
in energy

Analyzing the L
-
C Circuit

Total energy in the circuit:

Differentiate :

N o change
in energy

Analyzing the L
-
C Circuit

Total energy in the circuit:

Differentiate :

N o change
in energy

The charge sloshes back and

forth with frequency
w
㴠⡌䌩
-
1/2

LC Oscillations

Work out equation for LC circuit (loop rule)

Rewrite using i = dq/dt

w

(angular frequency) has dimensions of 1/t

Identical to equation of mass on spring

L

C

LC Oscillations (2)

Solution is same as mass on spring

oscillations

q
max

is the maximum charge on capacitor

is an unknown phase (depends on initial
conditions)

Calculate current: i = dq/dt

Thus both charge and current oscillate

Angular frequency
w
, frequency f =
w
/2
p

Period: T = 2
p
/
w

Plot Charge and Current vs t

Energy Oscillations

Total energy in circuit is conserved. Let’s
see why

(Multiply by i = dq/dt)

Equation of LC circuit

Use

U
L

+ U
C

= const

Oscillation of Energies

Energies can be written as (using
w
2

= 1/LC)

Conservation of energy:

Energy oscillates between capacitor
and
inductor

Endless oscillation between electrical and magnetic energy

Just like oscillation between potential energy and kinetic energy
for mass on spring

Plot Energies vs t

Sum

UNDRIVEN

RLC Circuit

Work out equation using loop rule

Rewrite using i = dq/dt

Solution slightly more complicated than LC
case

This is a damped oscillator (similar to
mechanical case)

Amplitude of oscillations falls exponentially

Charge and Current vs t in RLC Circuit

RLC Circuit (Energy)

Basic RLC equation

Multiply by i = dq/dt

Collect terms

(similar to LC circuit)

Total energy in circuit

decreases at rate of i
2
R

(dissipation of energy)

Energy in RLC Circuit

Sum

AC Circuits

Enormous impact of AC circuits

Power delivery

Tuners

Filters

Transformers

Basic components

R

L

C

Driving emf

AC Circuits and Forced
Oscillations

RLC + “driving” EMF with angular frequency
w
d

General solution for current is sum of two terms

“Transient”: Falls

exponentially & disappears

Constant amplitude

Ignore

AC Circuits and Forced
Oscillations

Assume steady state solution of form

I
m

is current amplitude

f

is phase

by which current “lags” the driving EMF

Must determine I
m

and
f

Plug in solution: differentiate & integrate sin(
w
t
-
f
)

Substitute

Current (2)

Expand sin & cos expressions

Collect sin
w
d
t &

cos
w
d
t

terms separately

These equations can be solved for I
m

and
f

(next slide)

High school trig!

cos
w
d
t terms

sin
w
d
t terms

Solve for
f

and I
m

in terms of

R, X
L
, X
C

and Z have dimensions of resistance

Let’s try to understand this solution using
“phasors”

Solution

for AC Current (3)

Inductive “reactance”

Capacitive “reactance”

Total “impedance”

REMEMBER Phasor Diagrams?

V
p

I
p

w
t

V
p

I
p

w
t

V
p

I
p

w
t

Resistor

Capacitor

Inductor

A phasor is an arrow whose length represents the amplitude of

an AC voltage or current.

The phasor rotates counterclockwise about the origin with the

angular frequency of the AC quantity.

Phasor diagrams are useful in solving complex AC circuits.

Reactance
-

Phasor Diagrams

V
p

I
p

w
t

V
p

I
p

w
t

V
p

I
p

w
t

Resistor

Capacitor

Inductor

“Impedance”

of an AC Circuit

R

L

C

~

The impedance, Z, of a circuit relates peak

current to peak voltage:

(Units: OHMS)

“Impedance”

of an AC Circuit

R

L

C

~

The impedance, Z, of a circuit relates peak

current to peak voltage:

(Units: OHMS)

(This is the AC equivalent of Ohm’s law.)

Impedance of an RLC Circuit

R

L

C

~

E

As in DC circuits, we can use the loop method:

E

-

V
R

-

V
C

-

V
L

= 0

I is same through all components.

Impedance of an RLC Circuit

R

L

C

~

E

As in DC circuits, we can use the loop method:

E

-

V
R

-

V
C

-

V
L

= 0

I is same through all components.

BUT: Voltages have different PHASES

Phasors for a Series RLC Circuit

I
p

V
Rp

(V
Cp
-

V
Lp
)

V
P

f

V
Cp

V
Lp

Phasors for a Series RLC Circuit

By Pythagoras’ theorem:

(V
P

)
2

= [ (V
Rp

)
2

+ (V
Cp

-

V
Lp
)
2

]

I
p

V
Rp

(V
Cp
-

V
Lp
)

V
P

f

V
Cp

V
Lp

Phasors for a Series RLC Circuit

By Pythagoras’ theorem:

(V
P

)
2

= [ (V
Rp

)
2

+ (V
Cp

-

V
Lp
)
2

]

= I
p
2

R
2

+ (I
p

X
C

-

I
p

X
L
)

2

I
p

V
Rp

(V
Cp
-

V
Lp
)

V
P

f

V
Cp

V
Lp

Impedance of an RLC Circuit

Solve for the current:

R

L

C

~

Impedance of an RLC Circuit

Solve for the current:

Impedance:

R

L

C

~

The circuit hits resonance when 1/
w
C
-
w
L=0:
w
r
=1/

When this happens the capacitor and inductor cancel each other

and the circuit behaves purely resistively: I
P
=V
P
/R.

Impedance of an RLC Circuit

The current’s magnitude depends on

the driving frequency. When Z is a

minimum, the current is a maximum.

This happens at a resonance frequency:

w

The current dies away

at both low and high

frequencies.

I
P

0

1

0

2

1

0

3

1

0

4

1

0

5

R

=

1

0

0

W

R

=

1

0

W

w
r

L=1mH

C=10
m
F

Phase in an RLC Circuit

I
p

V
Rp

(V
Cp
-

V
Lp
)

V
P

f

V
Cp

V
Lp

We can also find the phase:

tan
f

= (V
Cp

-

V
Lp
)/ V
Rp

or;

tan
f

= (X
C
-
X
L
)/R
.

or

tan
f

= (1/
w
C
-

w
L) / R

Phase in an RLC Circuit

At resonance the phase goes to zero (when the circuit becomes

purely resistive, the current and voltage are in phase).

I
p

V
Rp

(V
Cp
-

V
Lp
)

V
P

f

V
Cp

V
Lp

We can also find the phase:

tan
f

= (V
Cp

-

V
Lp
)/ V
Rp

or;

tan
f

= (X
C
-
X
L
)/R
.

or

tan
f

= (1/
w
C
-

w
L) / R

More generally, in terms of impedance:

cos
f =

R/Z

Power in an AC Circuit

V(t) = V
P

sin (
w

=
I(t) = I
P

sin (
w

=

=
I
P
V
P

sin
2
(
w

=

=
=
twice as fast.

V

w
t

p

2p

I

w
t

p

2p

P

f
= 0

(This is for a purely

resistive

circuit.)

The power is P=IV. Since both I and V vary in time, so

does the power: P is a function of time.

Power in an AC Circuit

Use,
V = V
P

sin (
w
t) and I = I
P

sin (
w
t+
f
) :

P(t) = I
p
V
p
sin(
w
t) sin (
w
t+
f
)

This wiggles in time, usually very fast. What we usually

care about is the time average of this:

(T=1/f )

Power in an AC Circuit

Now:

Power in an AC Circuit

Now:

Power in an AC Circuit

Use:

and:

So

Now:

Power in an AC Circuit

Use:

and:

So

Now:

which we usually write as

Power in an AC Circuit

f

goes from
-
90
0

to 90
0
, so the average power is positive)

cos(
f)

is called the
power factor
.

For a purely resistive circuit the power factor is 1.

When R=0, cos(
f
)=0 (energy is traded but not dissipated).

Usually the power factor depends on frequency.

Power in an AC Circuit

What if
f

is not zero?

I

V

P

Here I and V are 90
0

out of phase. (
f=
90
0
)

(It is purely
reactive
)

The
time average

of

P is zero.

w
t

STOP!

Transformers

Transformers use mutual inductance to change voltages:

Primary

Secondary

N
2

turns

V
1

V
2

N
1

turns

Iron Core

Power is conserved, though:

(if 100% efficient.)

Transformers & Power Transmission

20,000 turns

V
1
=110V

V
2
=20kV

110 turns

Transformers can be used to “step up” and “step

down” voltages for power transmission.

Power

=I
1

V
1

Power

=I
2

V
2

We use high voltage (e.g. 365 kV) to transmit electrical

power over long distances.

Why

do we want to do this?

Transformers & Power Transmission

20,000 turns

V
1
=110V

V
2
=20kV

110 turns

Transformers can be used to “step up” and “step down”
voltages, for power transmission and other applications.

Power

=I
1

V
1

Power

=I
2

V
2

We use high voltage (e.g. 365 kV) to transmit electrical

power over long distances.

Why

do we want to do this?
P = I
2
R

(P = power dissipation in the line
-

I is smaller at high voltages)