ECE201 Lect-5 1

Single-Node-Pair Circuits (2.4); Sinusoids (7.1);

Dr. S. M. Goodnick

September 5, 2002

ECE201 Lect-5 2

Example: 3 Light Bulbs in Parallel

How do we find I1, I2, and I3?

I R2 V

+

–

R1

I1 I2

R3

I3

ECE201 Lect-5 3

Apply KCL at the Top Node

I= I1 + I2 + I3

11 R

VI

22 R

VI

33 R

VI

ECE201 Lect-5 4

Solve for V

321321

111

RRRV

R

V

R

V

R

VI

321

1111

RRR

IV

ECE201 Lect-5 5

Req

321

1111

RRR

Req

iMpar RRRRR

11111

21

Which is the familiar equation for parallel resistors:

ECE201 Lect-5 6

Current Divider

• This leads to a current divider equation for three or more parallel resistors.

• For 2 parallel resistors, it reduces to a simple form.

• Note this equation’s similarity to the voltage divider equation.

j

parSR R

RII

j

ECE201 Lect-5 7

Is2 VR1 R2

+

–

I1 I2

Example: More Than One Source

How do we find I1 or I2?

Is1

ECE201 Lect-5 8

Apply KCL at the Top Node

I1 + I2 = Is1 - Is2

212121

11

RRV

R

V

R

VII ss

21

2121 RR

RRIIV ss

ECE201 Lect-5 9

Multiple Current Sources

• We find an equivalent current source by algebraically summing current sources.

• As before, we find an equivalent resistance.

• We find V as equivalent I times equivalent R.

• We then find any necessary currents using Ohm’s law.

ECE201 Lect-5 10

In General: Current Division

Consider N resistors in parallel:

Special Case (2 resistors in parallel)

iNpar

j

parSR

RRRRR

R

Rtiti

kj

11111

)()(

21

21

2)()(1 RR

Rtiti SR

ECE201 Lect-5 11

Class Examples

• Learning Extension E2.11

ECE201 Lect-5 12

Sinusoids: Introduction

• Any steady-state voltage or current in a linear circuit with a sinusoidal source is a sinusoid.– This is a consequence of the nature of

particular solutions for sinusoidal forcing functions.

– All steady-state voltages and currents have the same frequency as the source.

ECE201 Lect-5 13

Introduction (cont.)

• In order to find a steady-state voltage or current, all we need to know is its magnitude and its phase relative to the source (we already know its frequency).

• Usually, an AC steady-state voltage or current is given by the particular solution to a differential equation.

ECE201 Lect-5 14

The Good News!

• We do not have to find this differential equation from the circuit, nor do we have to solve it.

• Instead, we use the concepts of phasors and complex impedances.

• Phasors and complex impedances convert problems involving differential equations into simple circuit analysis problems.

ECE201 Lect-5 15

Phasors

• A phasor is a complex number that represents the magnitude and phase of a sinusoidal voltage or current.

• Remember, for AC steady-state analysis, this is all we need---we already know the frequency of any voltage or current.

ECE201 Lect-5 16

Complex Impedance

• Complex impedance describes the relationship between the voltage across an element (expressed as a phasor) and the current through the element (expressed as a phasor).

• Impedance is a complex number.

• Impedance depends on frequency.

ECE201 Lect-5 17

Complex Impedance (cont.)

• Phasors and complex impedance allow us to use Ohm’s law with complex numbers to compute current from voltage, and voltage from current.

ECE201 Lect-5 18

Sinusoids

• Period: T– Time necessary to go through one cycle

• Frequency: f = 1/T– Cycles per second (Hz)

• Angular frequency (rads/sec): = 2 f

• Amplitude: VM

tVtv M cos)(

ECE201 Lect-5 19

Example

What is the amplitude, period, frequency, and angular (radian) frequency of this sinusoid?

-8

-6

-4

-2

0

2

4

6

8

0 0.01 0.02 0.03 0.04 0.05

ECE201 Lect-5 20

Phase

-8

-6

-4

-2

0

2

4

6

8

0 0.01 0.02 0.03 0.04 0.05

ECE201 Lect-5 21

Leading and Lagging Phase

x1(t) leads x2(t) by -x2(t) lags x1(t) by -

On the preceding plot, which signals lead and which signals lag?

tXtx M cos)(11

tXtx M cos)(22

ECE201 Lect-5 22

Class Examples

• Learning Extension E7.1

• Learning Extension E7.2