l23 power and resonance in ac circuits

8/20/2019 L23 Power and Resonance in AC Circuits

1/17

Lecture 23. Impedance, Resonance in R-C-L Circuits

1

Preparation for the Final Exam

(a) Start earlier!(b) Review the concepts (lectures + textbook) and prepare your equation

sheet. Think how you can use every equation on your sheet, what types of

problems can be solved with these equations.

(c) Work on practice exams.

(d) Review all HW and Iclicker questions.

(e) Go over the end-of-chapter problems (you don’t need to solve them,

just check that you know how to approach them).

At the Exam(a) Make sure you understand the problem, read the problem formulation

carefully. Make a drawing!!! If you remain uncertain raise your hand and

ask the proctors.

(b) Get the units right. It is easy to eliminate the answers with wrong units.

This applies to formulas too.


2/17

Reactance (recap)

2

= =

Resistor

Capacitor

Inductor

=

=1

=

=

=

AC (cos + ) driven circuits!


3/17

Impedance

3

=

Impedance is a measure of how much the circuit

impedes the flow of current. The impedance is a complex

number (time-independent phasor), it relates time-

dependent phasors V(t) and I(t).

=

RLC

Reactances: = =1

=

Impedances: = =

1

=

= =

all terms are real

=

−

is the reference

phasor

=

−

≡


4/17

C circuit

Can we plug a 1-µF capacitor into a wall outlet

( = 2 ∙ 60

, = 120) if the circuit breakers

can take 15A ?

= =1

=

1

2 ∙ 60 ∙ 1 ∙ 10−Ω = 2650Ω

=

=

120

2650Ω = 45

This current is sufficiently small. Theprimary concern is the voltage rating

of the capacitor, which should be

around 200V.

4

= =

Voltage = =

Current (reference phasor)


5/17

L circuit

What happens when we plug a 1-H inductor into a wall

outlet?

= = = 2 ∙ 60 ∙ 1Ω = 377Ω

=

= 120377Ω

= 0.32

Again, the current won’t blow a circuit

breakers. The inductor must be designed tocarry 0.32A without overheating or

saturating the iron core.

5

= =

Voltage

Current (reference phasor)

= =


6/17

Series R-C circuit

6

= = ()

=

+

=

+

1

= + =

= + =

=

=

−

=

1


7/17

R-C circuits: Example

7

() = () =

−

= 1

tan =

1

= 1

1 ∙ 10 ∙ 100 ∙ 0.2 ∙ 10− = 50

1

Note that is negative (as it should be for the RC circuits).

=

=


8/17

Low-Pass Filter

8

=

Goal: to suppress high-frequency ( > )

components in the spectrum of a signal.

=

=

+

=

1

+

1

=

1

+ 1=

1

+ 1

=

= +

=

=1

+ 1

≪1

1

≫1

1

Cutoff frequency: = 2 =1

We want to suppress the high-frequency ( > 10)

components in the output of an audio amplifier with the

output resistance 100 Ω. What capacitance do you need?

=1

2

=1

210

∙ 100

= 160

Output power:= two

times

=

= +


9/17

Series R-L-C Circuits

9

= + = + 1

For R, C , and L in series:

() = () +

1

= ∙ ∗ = + 1

= = +

1

?1


10/17

Series R-L-C circuits: Example 1

10

=

= = +

1

= 2.2 40 + 80 110 = 110

= 2.2 ∙ 80 = 176

=

= 2.2 ∙ 110 = 242

= 2.2 ∙ 40 = 88


11/17

Series R-L-C circuits: Example 2

11

= ∙ cos = 151 ∙ 0.446 ∙ 0.72 = 48.6

An R-L-C series circuit with an inductance of 0.119H , a resistance of

244 Ω, and a capacitance of 7.27 µF carries an rms current of 0.446A

with a frequency of 391Hz .

= + = + 1

=

+ 1

= 339Ω

tan =

1

=

2455 ∙ 0.119 2455 ∙ 7.27 ∙ 10− −

244 ≈ 0.97

= 2455 /

arctan 0.97 ≈ 0.77

cos 0.77 = 0.72

3. What is the rms voltage of the source?

4. What average power is delivered by the source?

- average rate at which electrical energy is converted to thermal energy in the resistor

2. What is the phase angle?

- power factor for this circuit

1. What is the impedance of the circuit?

= ∙ = 0.446 ∙ 339Ω = 151


12/17

Parallel R-L-C Circuit: Example

12

= ∙ cos

=

=

1

+

1

=10

2

1

0.25 + 6 3 =

50

2

1

=1

+ 1

1

=

1

+

1

=

1

1

+ 1

=

1

1

1

+ 1

tan =

1

1/ =

3

4 cos =

1

1 + =

4

5

= ∙ cos =10

2∙

50

2∙

4

5 = 200


13/17

Series Resonance in the R-L-C circuits

13

= + = + 1

For R, C , and L in series:

= ∙ ∗

=

+

1

=

=

+ 1

=

= 1

- resonancefrequency

At = minimum (real) impedance, max current.

> < =

Note that at =, and can be greater than .

Resonancecondition:


14/17

Parallel Resonance in the R-L-C circuits

16

1

=

1

+

1

+

=

1

1

+

1

=

1

+

1

1

=

1

+

1

=

=

1

+

1

min at =

=1

At the resonance frequency

R = 1Ω, C = 1F, L = 1H, and V = 1V

is at its minimum

→ 0 is a “short”

→ ∞ is a “short”

Note that at =, and can be

greater than .


15/17

Transformer

17

Φ - the flux

per turn

ℇ = Φ

ℇ =

Φ

ℇ

ℇ=

For an ideal transformer

( = = 0):

=

Energy conservation: =

=

1

=

- as if the source had been

connected directly to a resistance

“impedance transformation”

=

Using mutual inductance = :

≡Φ

=Φ

= Φ

ℇ =


16/17

Example

18

ℇ =

=

Sloppy formulation

=

=0.4 ∙ cos 377 12

377 ∙ 5 ∙ cos 377 = 2.55

ℇ =

= 6 = 10 ∙ 10− ∙ 6 ∙ 3

= 0.18


17/17

19

Next time: Lecture 24. Electromagnetic Waves,

§§ 32.1 - 4

l23 power and resonance in ac circuits

Documents