Lect3 EEE 202 1

Voltage and Current Division; Superposition

Dr. Holbert

January 23, 2008

Lect3 EEE 202 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

+–

VS

R

R

R

I

Lect3 EEE 202 3

Solve for I

• 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

+–

VS

R

R

R

I + –

I R

+

–

I R

I R

+

–

N Total Resistors

IR + IR + … + IR – VS = 0

I = VS / (N R)

Lect3 EEE 202 4

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

Lect3 EEE 202 5

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

Lect3 EEE 202 6

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)()(

Lect3 EEE 202 7

Two Resistors in Parallel

How do we find I1 and I2?

I R1 R2 V

+

–

I1 I2

Lect3 EEE 202 8

Apply KCL with Ohm’s Law

212121

11

RRV

R

V

R

VIII

I R1 R2 V

+

–

I1 I2

21

21

21

111

RR

RRI

RR

IV

Lect3 EEE 202 9

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

Lect3 EEE 202 10

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

Lect3 EEE 202 11

Three Resistors in Parallel

I= I1 + I2 + I3

11 R

VI

22 R

VI

33 R

VI

I R2 V

+

–

R1

I1 I2

R3

I3

Lect3 EEE 202 12

Solve for V

321321

111

RRRV

R

V

R

V

R

VI

eqRI

RRR

IV

321

1111

Lect3 EEE 202 13

Equivalent Resistance (Req)

321

1111

RRR

Req

iMpar RRRRR

11111

21

Which is the familiar equation for parallel resistors:

Lect3 EEE 202 14

Current Divider

• This leads to a current divider equation for multiple 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

Lect3 EEE 202 15

Example: More Than One Source

How do we find I1 or I2?

Is1 Is2 VR1 R2

+

–

I1 I2

Lect3 EEE 202 16

Apply KCL at the Top Node

21212121

11

RRV

R

V

R

VIIII ss

21

2121 RR

RRIIV ss

Is1 Is2 VR1 R2

+

–

I1 I2

Lect3 EEE 202 17

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

Lect3 EEE 202 18

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

Lect3 EEE 202 19

Superposition

“In any linear circuit containing multiple independent sources, the current or voltage at any point in the circuit may be calculated as the algebraic sum of the individual contributions of each source acting alone.”

Lect3 EEE 202 20

How to Apply Superposition

• To find the contribution due to an individual independent source, zero out the other independent sources in the circuit

– Voltage source short circuit

– Current source open circuit

• Solve the resulting circuit using your favorite technique(s)

Lect3 EEE 202 21

Superposition of Summing Circuit

+

–

Vout

1k

1k

1k

V1 V2

+–

+–

+

–

V’out

1k

1k

1k

V1

+

–

V’’out

1k

1k

1k

V2++–

+–

Lect3 EEE 202 22

Use of Superposition

V’out = V1/3

V’’out = V2/3

Vout = V’out + V’’out = V1/3 + V2/3

+

–

V’out

1k

1k

1k

V1

+

–

V’’out

1k

1k

1k

V2++–

+–

Lect3 EEE 202 23

Superposition Procedure

1. For each independent voltage and current source (repeat the following): a) Replace the other independent voltage sources with

a short circuit (i.e., V = 0). b) Replace the other independent current sources with

an open circuit (i.e., I = 0). Note: Dependent sources are not changed!

c) Calculate the contribution of this particular voltage or current source to the desired output parameter.

2. Algebraically sum the individual contributions (current and/or voltage) from each independent source.

Lect3 EEE 202 24

Class Examples

• Drill Problems P2-2 & P2-4, P2-7, P2-1 & P2-3