parallel circuits 2

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ECR255 - 1 – Seneca College – School of Information and Communications Technology G VandeBelt 112 Slide 1 ECR255 Parallel AC Circuits ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ____________________________________________________________________

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Page 1: Parallel Circuits 2

ECR255

- 1 – Seneca College – School of Information and Communications Technology G VandeBelt 112

Slide 1

ECR255

Parallel AC Circuits

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Page 2: Parallel Circuits 2

ECR255

- 2 – Seneca College – School of Information and Communications Technology G VandeBelt 112

Slide 2

Find the Elements

Consider an unknown circuit in a box connected as shown

If total power P = 4 W with a leading power factor, what is

the simplest possible parallel circuit in the box if the

frequency is 1 kHz?

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Page 3: Parallel Circuits 2

ECR255

- 3 – Seneca College – School of Information and Communications Technology G VandeBelt 112

Slide 3

Find the Elements

P = E I cos θ

cos θ = P / E I = 4 W / 40 V x 150 mA = 0.6667

θ = 48.19o current leads voltage

I = 150 mA /48.19o

YT = I / E = 150 mA /48.19o / 40 V /0o = 3.75 mS /48.19o

= 2.50 mS + j 2.795 mS

= 1/ R + Gx + j Bx

Gx = 2.50 mS - 1.0 mS = 1.50 mS Rx = 667 Ω

Bx = 2.795 mS = 2π f Cx

Cx = Bx / 2π f = 2.795 mS / 2π x 103 = 445 nF3

P = 4 W leading - capacitivef = 1 kHz

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Page 4: Parallel Circuits 2

ECR255

- 4 – Seneca College – School of Information and Communications Technology G VandeBelt 112

Slide 4

Current Divider Rule

The basic format for the current divider rule in

ac circuits is exactly the same as that for dc

circuits; that is, for two parallel branches with

impedances Z1 and Z2 :

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Page 5: Parallel Circuits 2

ECR255

- 5 – Seneca College – School of Information and Communications Technology G VandeBelt 112

Slide 5

Example

Calculate the currents I1 I2 in phasor form using the current

divider rule

ZR = 22 Ω /0o

ZL = (XL1 + XL2) /90o = 80 Ω /90o

ZR + ZL = 22 Ω + j 80 Ω

ZR + ZL = 82.97 Ω /74.62o

I1 = I ZL / (ZR + ZL )

I1 = 20 A /40o x 80 Ω /90o / 82.97 Ω /74.62o = 19.3 A /55.38o

I2 = I ZR / (ZR + ZL )

I3 = 20 A /40o x 22 Ω /0o / 82.97 Ω /74.62o = 5.30 A /34.62o

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XL1

XL2

R

22 Ω

60 Ω

20 Ω

I2I1

I = 20 A /40o

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Page 6: Parallel Circuits 2

ECR255

- 6 – Seneca College – School of Information and Communications Technology G VandeBelt 112

Slide 6

Examples

For the circuit

a) Find the total admittance in polar form

b) Draw the admittance diagram

c) Calculate the capacitance and inductance of the devices

d) Find voltage E and currents IR IC IL in polar form

e) Draw phasor diagram of IS IR IC IL and E

f) Verify Kirchhoff’s current law at one node.

g) Find average power delivered to the load

h) Find power factor. Is it leading or lagging?

i) Find sinusoidal expressions for currents and voltage

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1.2 Ω

RX

L

XC

is

3 sin(377t+60o) 2 Ω 5 Ω

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Page 7: Parallel Circuits 2

ECR255

- 7 – Seneca College – School of Information and Communications Technology G VandeBelt 112

Slide 7

Frequency Response of Parallel Elements

As the frequency increases, XL will increase

until it equals the impedance of the resistor R.

For frequencies less than ƒ2: X

L< R. For

frequencies greater that ƒ2: X

L> R.

The magnitude and angle of the total

impedance can now be found at any

frequency of interest by substituting the

following equations:

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Page 8: Parallel Circuits 2

ECR255

- 8 – Seneca College – School of Information and Communications Technology G VandeBelt 112

Slide 8

Frequency Response of Parallel Elements

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