AC electric circuits 1.More difficult than DC circuits 2. Much more difficult than DC circuits 3....

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AC electric circuits 1.More difficult than DC circuits 2. Much more difficult than DC circuits 3. You can do it!

Transcript of AC electric circuits 1.More difficult than DC circuits 2. Much more difficult than DC circuits 3....

AC electric circuits

1.More difficult than DC circuits

2. Much more difficult than DC circuits

3. You can do it!

DC: Direct current

AC: Alternating current: amplitude and direction vary with time.

There are many situations where working with AC is more effective. For example,AC signal can be stepped up or down using a transformer. (Amp—Speakers)

AC voltage over time

To measure the voltage:Peak voltage dependson peak shape.AC is not necessarily sine waves

AC signal of various forms and looks

RMS: Root mean squarevalue

For a sine wave

V =Vm sinωt

Vrms =Vm2

AC Phase

Two signals out of step with each other

A leads B by 45 degrees

AC resistor circuits

Voltage and current in phase€

ET = E0 sinωt

I =ETR=E0Rsinωt

Power = I V, always positive.Resistors always consumes energy

AC Inductor circuits

EL = Ldi

dtI = IL = I0 sinωt

EL = LωI0 cosωt

In AC circuits, it is not enough just to state the magnitude of a current or a voltage

We need to know the timing of the signals.A useful way to illustrate the timing, is the Use of phasor diagrams.

IL

EL90 degreesAC Inductor circuits

AC resistor circuitsIR

ER

Current lags voltage by 90 degrees

EL = Ldi

dtI = IL = I0 sinωt

EL = LωI0 cosωt

The instantaneous power can be negative! i.e. The inductor absorbs power, as well as releases power.

When I and V are both positive, P is positive.When I and V are both negative, P is also positive.Since I and V are 90 degrees out of phase, There are times when one of them is positive,the other is negative, thus P is negative.

When P is positive, inductor absorbs power, when P is negative, it releases power.

For a pure inductive circuit, the average power consumption is

Zero

Compare with R, which passively consumespower, L is a reactive load.

EL = LωI0 cosωt

Voltage amplitude across the inductor is proportional to the angular frequency.

The “reactance” of an inductor:

XL = Lω

High w, faster change in current,Hence higher voltage.

Unit ?W

VL = LωI0 cosωtor

The reactance has two important components:1) Its magnitude:Which relates the amplitude of the current to that of the voltage.

2) It controls the phase difference between the current and the voltage.

XL = Lω

Vp = IpLω

Reactance of L: XL=3.7699W

I =10V

3.7699Ω= 2.6526A

reac tance =Voltage

Current=Voltage∠900

Current∠00

Reac tance = 3.7699Ω∠900

Reac tance = 3.7699Ω∠900

EL = LωI0 cosωt

This is the amplitude of current

XL has a magnitude Lw, and carries a phase 90 degrees

Compare with resistance,The inductance is an active component.The amplitude and the phase of the current are controlled by the property of L

XL = 0 + j3.7699Ω

We use j to indicate the phase of the reactance

reactanceR

Series resistor-inductor circuits

Z:Impedance

Z = R + jXL

Z has a magnitude as well as a phase

The magnitude of Z :

R2 + XL2

The phase of Z is f :

tanφ =XLR

Impedance Z= R+jXL

Ohm’s law for AC circuits

V = IZ

I =V

Z

For the circuit

I =V

Z

We choose V has zero phase

I =V

Z=

10V∠00

6.26Ω∠37.0160=1.597A∠− 37.0160

Current lags voltage by 37 degrees, less than 90 degrees for the pure inductance circuit. This is because of the effect of the resistor. RL time constant depends on R.

RL impedance phasor diagram

Parallel R-L circuits

ZT =1

1

Z1+1

Z2+1

Z3+ ...

1

Z1=1

R= 0.2

1

Z2=1

jωL=

1

j2π ×60 ×10−2=

1

j3.77

1

Z1+1

Z2= 0.2 +

1

j3.77=1

Z

I =V

Z=V

Z=10 × (0.2 +

1

j3.77) =10 × (0.2 −

j

3.77) = 2 − j2.65

I =V

Z= 3.31< −530

or

AC capacitive circuits

Ic =CdV

dtif

V =V0 sinωt, then

Ic =CdV

dt=V0Cωcosωt

Ic =CdV

dt=V0ωCcosωt = I0 cosωt

XC =V0I0=1

ωC

Capacitive reactance

The phase of XC is -90 degrees.

XC = 26.53Ω

I =10V

26.53Ω= 0.38

opposition =Voltage

Current=Voltage∠00

Current∠900=10V∠00

0.378∠900= 26.53Ω∠−900

For capacitance C

Zc = −j

ωC=1

jωC

Series RC

ZT = 5 +1

j2π ×60 ×10−4= 5 − j2.63

Parallel RC

Series RLC

ZT = 250 + j245.04 − j1768.4 = 250 − j1523

In general, for series RLC circuits:

ZT = R + jωL +1

jωC= R + jωL −

j

ωC= R + j(ωL −

1

ωC)

R is independent of frequency.The imaginary part is frequency dependent,and it equals to zero when

ωL = 1

ωC

ω =1

LC

ω =1

LC

ZT = R

When

The impedance is purely resistive

The impedance is minimumSo the current flowing in the circuit reaches maximumUnder such a condition, we say the circuitIs in resonance.

ω =1

LC

is the resonance angular frequency for thecircuit.

ZT = R + jωL +1

jωC= R + jωL −

j

ωC= R + j(ωL −

1

ωC)

When the circuit is not in resonance

The magnitude of the ZT:

tanφ =ωL −

1

ωCR€

R2 + (ωL −1

ωC)2

Resonance