ece 1100: introduction to electrical and computer engineering voltage and current divider rules...
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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 .