ece 1100: introduction to electrical and computer engineering voltage and current divider rules...
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![Page 1: ECE 1100: Introduction to Electrical and Computer Engineering Voltage and Current Divider Rules Notes 19 Spring 2011 Wanda Wosik Associate Professor, ECE](https://reader036.vdocuments.net/reader036/viewer/2022062315/5697bff91a28abf838cbff43/html5/thumbnails/1.jpg)
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
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Voltage Divider RuleVoltage Divider Rule
+
-vs (t)
R1
R2
+
-
v2 (t)
Find v2 (t)
Note: there is an open circuit at the output terminals.
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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
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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
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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 :
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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.)
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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.)
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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.)
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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.)
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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!
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Current Divider RuleCurrent Divider Rule
vs (t)
+
- R1 R2
i (t)
i1 (t) i2 (t)
Find i2 (t)
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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)
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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)
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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
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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
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+
-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.)
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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.)
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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.)
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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 .