Voltage and Current
Division; Superposition
Dr. Holbert
January 23, 2008
Lect3
EEE 202
1
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
Lect3
EEE 202
I
R
R
+
–
VS
R
2
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
Lect3
I + IR
–
+
R
R
+
–
IR
–
VS
N Total
Resistors
+
R
IR
–
IR + IR + … + IR – VS = 0
I = VS / (N  R)
EEE 202
3
In General: Single Loop
• The current i(t) is:
 VSi sum of voltage sources
i (t ) 

 Rj
sum of resistances
• This approach works for any single loop
circuit with voltage sources and resistors
• Resistors in series
Rseries  R1  R2    RN   R j
Lect3
EEE 202
4
Voltage Division
Consider two resistors in series with a
voltage v(t) across them:
+
+
R1
R2
Lect3
v2(t)
R2
v2 (t )  v(t )
R1  R2
–
+
v(t)
–
v1(t)
R1
v1 (t )  v(t )
R1  R2
–
EEE 202
5
In General: Voltage Division
• Consider N resistors in series:
Ri
VRi (t )  VSk (t )
 Rj
• Source voltage(s) are divided between the
resistors in direct proportion to their
resistances
Lect3
EEE 202
6
Two Resistors in Parallel
I
I1
R1
I2
R2
+
V
–
How do we find I1 and I2?
Lect3
EEE 202
7
Apply KCL with Ohm’s Law
1
V V
1 
I  I1  I 2  
 V   
R1 R2
 R1 R2 
I1
R1
I
I2
R2
+
V
–
1
Lect3
R1 R2
V I
I
1
1
R

R
1
2

R1 R2
EEE 202
8
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:
R1R2
Req 
R1  R2
Definition: Parallel - the elements share the
same two end nodes
Lect3
EEE 202
9
Now to find I1
R1 R2
I
V
R2
R1  R2
I1 

I
R1
R1
R1  R2
• This is the current divider formula
• It tells us how to divide the current through
parallel resistors
Lect3
EEE 202
10
Three Resistors in Parallel
I= I1 + I2 + I3
I1
R1
I
I2
R2
I3
R3
+
V
–
V
I1 
R1
Lect3
V
I2 
R2
EEE 202
V
I3 
R3
11
Solve for V
1
V V V
1
1
I 

 V  
 
R1 R2 R3
 R1 R2 R3 
1
V I
 I Req
1
1
1


R1 R2 R3
Lect3
EEE 202
12
Equivalent Resistance (Req)
1
Req 
1
1
1


R1 R2 R3
Which is the familiar equation for parallel resistors:
1
1
1
1
1
 


R par R1 R2
RM
Ri
Lect3
EEE 202
13
Current Divider
• This leads to a current divider equation for
multiple parallel resistors
IRj  IS
R par
Rj
• For 2 parallel resistors, it reduces to a
simple form
• Note this equation’s similarity to the
voltage divider equation
Lect3
EEE 202
14
Example: More Than One
Source
Is1
I1
R1
Is2
I2
R2
+
V
–
How do we find I1 or I2?
Lect3
EEE 202
15
Apply KCL at the Top Node
I s1  I s 2
Is1
 1
V
V
1 

 I1  I 2 

 V  
R1 R2
 R1 R2 
I1
R1
Is2
I2
R2
+
V
–
R1 R2
V  I s1  I s 2 
R1  R2
Lect3
EEE 202
16
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
17
In General: Current Division
• Consider N resistors in parallel:
iR j (t )   iS k (t )
R par
Rj
1
1
1
1
1
 


R par R1 R2
RN
Ri
• Special Case (2 resistors in parallel)
R2
iR1 (t )  iS (t )
R1  R2
Lect3
EEE 202
18
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
19
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
20
Superposition of Summing Circuit
1kW
1kW
+
+
–
V1
Vout
1kW
+
–
V2
–
1kW
1kW
1kW
+
V1
+
–
V’out
+
1kW
+
–
Lect3
1kW
V’’out
1kW
+
–
V2
–
EEE 202
21
Use of Superposition
1kW
1kW
1kW
+
V1
+
–
V’out
1kW
+
1kW
+
–
V’’out
1kW
+
–
V2
–
V’out = V1/3
V’’out = V2/3
Vout = V’out + V’’out = V1/3 + V2/3
Lect3
EEE 202
22
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
23
Class Examples
• Drill Problems P2-2 & P2-4, P2-7, P2-1 &
P2-3
Lect3
EEE 202
24