ECE201 Lect-3 1

Single Loop Circuits (2.3); Single-Node-Pair Circuits (2.4)

Dr. Holbert

January 25, 2006

ECE201 Lect-3 2

Single Loop Circuit

• The same current flows through each element of the circuit---the elements are in series.

• We will consider circuits consisting of voltage sources and resistors.

ECE201 Lect-3 3

Example: Christmas Lights

+–120V

228

50 Bulbs Total

228

228

I

ECE201 Lect-3 4

Solve for I

• The same current I flows through the source and each light bulb-how do you know this?

• In terms of I, what is the voltage across each resistor? Make sure you get the polarity right!

• To solve for I, apply KVL around the loop.

ECE201 Lect-3 5

228I + 228I + … + 228I -120V = 0

I = 120V/(50 228) = 10.5mA

+–120V

228

228

228

I + –

228I

+

–

228I

228I

+

–

ECE201 Lect-3 6

Some Comments

• We can solve for the voltage across each light bulb:

V = IR = (10.5mA)(228 = 2.4V

• This circuit has one source and several resistors. The current is

Source voltage/Sum of resistances

(Recall that series resistances sum)

ECE201 Lect-3 7

In General: Single Loop

• The current i(t) is:

• This approach works for any single loop circuit with voltage sources and resistors.

• Resistors in series

sresistanceofsumsourcesvoltageofsum

RV

tij

Si )(

jNseries RRRRR 21

ECE201 Lect-3 8

Voltage Division

Consider two resistors in series with a voltage v(t) across them:

R1

R2

–

v1(t)

+

+

–

v2(t)

+

–

v(t)21

11 )()(

RR

Rtvtv

21

22 )()(

RR

Rtvtv

ECE201 Lect-3 9

In General: Voltage Division

Consider N resistors in series:

Source voltage(s) are divided between the resistors in direct proportion to their resistances

j

iSR R

RtVtV

ki)()(

ECE201 Lect-3 10

Class Examples

• Learning Extension E2.8

• Learning Extension E2.9

ECE201 Lect-3 11

Example: 2 Light Bulbs in Parallel

How do we find I1 and I2?

I R1 R2 V

+

–

I1 I2

ECE201 Lect-3 12

Apply KCL at the Top Node

I= I1 + I2

Ohm’s Law:1

1 R

VI

22 R

VI

I R1 R2 V

+

–

I1 I2

ECE201 Lect-3 13

Solve for V

212121

11

RRV

R

V

R

VIII

21

21

21

111

RR

RRI

RR

IV

Rearrange

ECE201 Lect-3 14

Equivalent Resistance

If we wish to replace the two parallel resistors with a single resistor whose voltage-current relationship is the same, the equivalent resistor has a value of:

Definition: Parallel - the elements share the same two end nodes

21

21

RR

RRReq

ECE201 Lect-3 15

21

2

1

21

21

11 RR

RI

R

RRRR

I

R

VI

Now to find I1

• This is the current divider formula.

• It tells us how to divide the current through parallel resistors.

ECE201 Lect-3 16

Example: 3 Light Bulbs in Parallel

How do we find I1, I2, and I3?

I R2 V

+

–

R1

I1 I2

R3

I3

ECE201 Lect-3 17

Apply KCL at the Top Node

I= I1 + I2 + I3

11 R

VI

22 R

VI

33 R

VI

ECE201 Lect-3 18

Solve for V

321321

111

RRRV

R

V

R

V

R

VI

321

1111

RRR

IV

ECE201 Lect-3 19

Req

321

1111

RRR

Req

iMpar RRRRR

11111

21

Which is the familiar equation for parallel resistors:

ECE201 Lect-3 20

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-3 21

Is2 VR1 R2

+

–

I1 I2

Example: More Than One Source

How do we find I1 or I2?

Is1

ECE201 Lect-3 22

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-3 23

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-3 24

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-3 25

Class Examples

• Learning Extension E2.10

• Learning Extension E2.11