Alternating Current CircuitsAlternating Current Circuits

Chapter 33Chapter 33

(continued)(continued)

Phasor Diagrams

VRp

Ipt

VCp

Ip

t

VLp Ip

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 current or voltage.

• 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 current or voltage.

Impedance in AC Circuits

The impedance Z of a circuit or circuit element relates peak current to peak voltage:

€

Ip =Vp

Z(Units: Ohms)

V

R

~ C

L

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

Phasor Diagrams

Circuit element Impedance Amplitude Phase

Resistor R VR= IP R I, V in phase

Capacitor Xc=1/C VC=IP Xc I leads V by 90°

Inductor XL=L VL=IP Xc I lags V by 90°

VRp

Ipt

VCp

Ip

t

VLp Ip

t

Resistor Capacitor Inductor

RLC Circuit

Use the loop method:

V - VR - VC - VL = 0

I is same through all components.

V

R

~ C

L

BUT: Voltages have different PHASES

they add as PHASORS.

RLC Circuit

Ip

VRp

(VCp- VLp)

VP

VCp

VLp

RLC Circuit

By Pythagoras’s theorem:

(VP )2 = [ (VRp )2 + (VCp - VLp)2 ]

= Ip2 R2 + (Ip XC - Ip XL)

2

Ip

VRp

(VCp- VLp)

VP

VCp

VLp

RLC Circuit

Solve for the current:

€

Ip =Vp

R2 + (Xc − XL )2=

Vp

Z

V

R

~ C

L

RLC Circuit

Solve for the current:

Impedance: €

Ip =Vp

R2 + (Xc − XL )2=

Vp

Z

€

Z = R2 +1

ωC−ωL

⎛

⎝ ⎜

⎞

⎠ ⎟2

V

R

~ C

L

The circuit hits resonance when 1/C-L=0: r=1/When this happens the capacitor and inductor cancel each otherand the circuit behaves purely resistively: IP=VP/R.

RLC Circuit

€

Ip =Vp

Z

Z = R2 +1

ωC−ωL

⎛

⎝ ⎜

⎞

⎠ ⎟2

The current’s magnitude depends onthe driving frequency. When Z is aminimum, the current is a maximum.This happens at a resonance frequency:

LC

The current dies awayat both low and highfrequencies.

IP

01 0

21 0

31 0

41 0

5

R = 1 0 0

R = 1 0

r

L=1mHC=10F

Phase in an RLC Circuit

IpVRp

(VCp- VLp)

VP

VCp

VLp

We can also find the phase:

tan = (VCp - VLp)/ VRp

= (XC-XL)/R = (1/C - L) / R

Phase in an RLC Circuit

At resonance the phase goes to zero (when the circuit becomespurely resistive, the current and voltage are in phase).

IpVRp

(VCp- VLp)

VP

VCp

VLp

We can also find the phase:

tan = (VCp - VLp)/ VRp

= (XC-XL)/R = (1/C - L) / R

More generally, in terms of impedance:

cos R/Z

The power dissipated in an AC circuit 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 = VP sin (t) and I = IP sin (t+) :

P(t) = IpVpsin(t) sin (t+)

This wiggles in time, usually very fast. What we usually care about is the time average of this:

€

P =1

TP(t)dt

0

T

∫ (T=1/f )

Power in an AC Circuit

Now:

€

sin(ωt + φ) = sin(ωt)cosφ + cos(ωt)sinφ

Power in an AC Circuit

€

P(t) = IPVP sin(ωt)sin(ωt + φ)

= IPVP sin2(ωt)cosφ + sin(ωt)cos(ωt)sinφ

Now:

€

sin(ωt + φ) = sin(ωt)cosφ + cos(ωt)sinφ

Power in an AC Circuit

€

P(t) = IPVP sin(ωt)sin(ωt + φ)

= IPVP sin2(ωt)cosφ + sin(ωt)cos(ωt)sinφ

€

sin2(ωt) =1

2

sin(ωt)cos(ωt) = 0

Use:

and:

So

€

P =1

2IPVP cosφ

Now:

€

sin(ωt + φ) = sin(ωt)cosφ + cos(ωt)sinφ

Power in an AC Circuit

€

P(t) = IPVP sin(ωt)sin(ωt + φ)

= IPVP sin2(ωt)cosφ + sin(ωt)cos(ωt)sinφ

€

sin2(ωt) =1

2

sin(ωt)cos(ωt) = 0

Use:

and:

So

€

P =1

2IPVP cosφ

Now:

which we usually write as

€

P = IrmsVrms cosφ

€

sin(ωt + φ) = sin(ωt)cosφ + cos(ωt)sinφ

Power in an AC Circuit

€

P = IrmsVrms cosφ

goes from -900 to 900, so the average power is positive)

cos( is called the power factor.

For a purely resistive circuit the power factor is 1.When R=0, cos()=0 (energy is traded but not dissipated).

Usually the power factor depends on frequency, and usually 0<cos()<1.

Power in a purely resistive circuit

V(t) = VP sin (t)

I(t) = IP sin (t)

P(t) = IV = IP VP sin 2(t) Note this oscillates

twice as fast.

V

t

I

t

P

= 0

(This is for a purely resistive circuit.)

Power in a purely reactive circuit

€

P = IrmsVrms cosφ

The opposite limit is a purely reactive circuit, with R=0.

IV P This happens with an

LC circuit.

Then 900

The time average of P is zero.

t

Transformers

Transformers use mutual inductance to change voltages:

Primary(applied voltage)

Secondary(produced voltage)

Ns turns

VpVs

Np turns Iron Core

€

Vs =Ns

N p

Vp

Faraday’s law on the left:If the flux per turn is then Vp=Np(d/dt).

Faraday’s law on the right:The flux per turn is also , so Vs=Ns(d/dt).

Transformers

Transformers use mutual inductance to change voltages:

Primary(applied voltage)

Secondary(produced voltage)

Ns turns

VpVs

Np turns Iron Core

€

Vs =Ns

N p

Vp

In the ideal case, no power is dissipated in the transformer itself.

Then IpVp=IsVs

€

Is =N p

Ns

Ip

Transformers & Power Transmission

20,000 turns

V1=110V V2=20kV

110 turns

Transformers can be used to “step up” and “stepdown” voltages for power transmission.

Power=I1 V1

Power=I2 V2

We use high voltage (e.g. 365 kV) to transmit electricalpower over long distances.Why do we want to do this?

Transformers & Power Transmission

20,000 turns

V1=110V V2=20kV

110 turns

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

Power=I1 V1

Power=I2 V2

We use high voltage (e.g. 365 kV) to transmit electricalpower over long distances.

Why do we want to do this? P = I2R (P = power dissipation in the line - I is smaller at high voltages)