ECE 1100: Introduction toECE 1100: Introduction toElectrical and Computer EngineeringElectrical and Computer Engineering

Voltage and Current Divider Rules

Notes 19

Spring 2011

Wanda WosikAssociate Professor, ECE Dept.

Notes prepared by Dr. Jackson

Voltage Divider RuleVoltage Divider Rule

+

-vs (t)

R1

R2

+

-

v2 (t)

Find v2 (t)

Note: there is an open circuit at the output terminals.

Voltage Divider RuleVoltage Divider Rule

+

-vs (t)R1

R2

+- v2 (t)

i (t)

i (t)

2 2 2

1 2

sv tv R i t R

R R

22

1 2s

Rv t v t

R R

Voltage Divider RuleVoltage Divider Rule

Rule: The multiplying factor that gives the voltage across a resistor is the resistance of the resistor divided by sum of the resistances.

+

-vs (t)R1

R2+

- v2 (t)

22

1 2s

Rv t v t

R R

ExampleExample

a) When measured with an ideal voltmeter (infinite internal resistance),b) When measured with a digital multimeter (DMM) that has a 10 [M]

resistance.

+

-10 [V] R1

R2

+- V2

R1 = 100 [] R2 = 200 []

Find the voltage V2 :

R1 = 100 []

R2 = 200 []

(a)

2

20010 6 666666667[V]

100 200V .

2 6 666666667[V]V .

+

-10 [V] R1

R2+

- V2

Example (cont.)Example (cont.)

R1 = 100 []

R2 = 200 []

(b)

+

-10 [V] R1

R2+

-V2RDMM

RDMM = 10 [M]

Note that R2 and RDMM are in parallel, so we can combine them.

Example (cont.)Example (cont.)

R1 = 100 []

R2 = 200 []

(b)

RDMM = 10 [M]

6

6

200 10 10199 9960001 [Ω]

200 10 10eqR .

+

-10 [V] R1

R2+

-V2RDMM

Example (cont.)Example (cont.)

R1 = 100 []

R2 = 200 []

(b)

Req = 199.9960001 []

+

-10 [V] R1

Req

+

-V2

2

199 996000110 6 666622222[V]

100 199 9960001

.V .

.

2 6 666622222[V]V .

Example (cont.)Example (cont.)

R1 = 100 [M]

R2 = 200 [M]

Example (cont.)Example (cont.)

Try the same example again using:

+

-10 [V] R1

R2+

-

V2RDMM

Ideal DMM: 2 6 6667[V]V .

Actual DMM: 2 0 8696 [V]V . There is a huge amount of loading by the DMM!

Current Divider RuleCurrent Divider Rule

vs (t)

+

- R1 R2

i (t)

i1 (t) i2 (t)

Find i2 (t)

Current Divider RuleCurrent Divider Rule

1 22 2 2 2

1 2

1

1 2

s eq

R Ri G v t G i t R G i t

R R

Ri t

R R

vs (t)

+

- R1 R2

i (t)

i1 (t) i2 (t)

Current Divider RuleCurrent Divider Rule

12

1 2

Ri i t

R R

Rule: The multiplying factor that gives the current through a resistor is the opposite resistance divided by the sum of the two resistances.

vs (t)

+

- R1 R2

i (t)

i1 (t) i2 (t)

ExampleExampleFind the RMS voltage across the two appliances and the current coming out of the outlet. Also find the current through RL2 and the average power dissipated by RL2.

I2

+

-Vs =120 [V] (RMS)

Rs = 3 []

RL1

+

-

V2RL2

RL1 = 144 [] RL2 = 14.4 []

RL1 = 100 [W] light bulb RL2 = 1000 [W] hair dryer

Rs = resistance of house wiring

outlet

Example (cont.)Example (cont.)

1 2

1 2

13 0909 [Ω]eq L LL

L L

R RR .

R R

RL1 = 144 [] RL2 = 14.4 []

+

-Vs =120 [V]

(RMS)

Rs = 3 []

RL1

+

-

V2RL2

I2

+

-Vs =120 [V]

(RMS)

Rs = 3 []

RLeq = 13.0909 []

+

-

V2

2

13 0909120 97 627 [V]

13 0909 3

.V .

.

2 (RMS)97 627 [V]V .

Example (cont.)Example (cont.)

1207 4576 [A]

3 13 0909sI ..

Example (cont.)Example (cont.)

+

-Vs =120 [V]

(RMS)

Rs = 3 []

RLeq = 13.0909 []

+

-

V2

Is

(This is the current (RMS) coming out of the outlet.)

12

1 2

1447 4576

144 14 4L

sL L

RI I .

R R .

RL1 = 144 [] RL2 = 14.4 []

+

-Vs =120 [V]Rs = 3 []

RL1

+

-

V2RL2

Is = 7.4576 [A]

I2

2 (RMS)6 7796 [A]I .

Example (cont.)Example (cont.)

Example (cont.)Example (cont.)

2222

2

97 627661 89 [W]

14 4RMSAVE

absL

V .P .

R .

Note: If there was 120 [V] (RMS) across the hair dryer, we would have

2

2 1201000 [W]

14 4AVE

absP.

2 661 89 [W]AVEabsP .