Single Loop Circuits (2.3);
Single-Node-Pair Circuits (2.4)
Dr. Holbert
January 25, 2006
ECE201 Lect-3
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.
ECE201 Lect-3
2
Example: Christmas Lights
I
228W
228W
120V
50 Bulbs
Total
+
–
228W
ECE201 Lect-3
3
Solve for I
• The same current I flows through the source
and each light bulb-how do you know this?
• 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.
ECE201 Lect-3
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I + 228I –
+
228W
228W
120V
228I
–
+
–
+
228W
228I
–
228I + 228I + … + 228I -120V = 0
I = 120V/(50  228W) = 10.5mA
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Some Comments
• We can solve for the voltage across each
light bulb:
V = IR = (10.5mA)(228W) = 2.4V
• This circuit has one source and several
resistors. The current is
Source voltage/Sum of resistances
(Recall that series resistances sum)
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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
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Voltage Division
Consider two resistors in series with a voltage
v(t) across them:
+
+
R1
v2(t)
R2
v2 (t )  v(t )
R1  R2
–
+
v(t)
R2
–
v1(t)
R1
v1 (t )  v(t )
R1  R2
–
ECE201 Lect-3
8
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
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Class Examples
• Learning Extension E2.8
• Learning Extension E2.9
ECE201 Lect-3
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Example: 2 Light Bulbs in
Parallel
I
I1
R1
I2
R2
+
V
–
How do we find I1 and I2?
ECE201 Lect-3
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Apply KCL at the Top Node
I= I1 + I2
Ohm’s Law:
I
V
I1 
R1
I1
R1
V
I2 
R2
I2
R2
+
V
–
ECE201 Lect-3
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Solve for V
1
V V
1 
I  I1  I 2  
 V   
R1 R2
 R1 R2 
Rearrange
1
R1 R2
V I
I
1
1
R

R
1
2

R1 R2
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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:
R1 R2
Req 
R1  R2
Definition: Parallel - the elements share the
same two end nodes
ECE201 Lect-3
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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.
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Example: 3 Light Bulbs in
Parallel
I
I1
R1
I2
R2
I3
R3
+
V
–
How do we find I1, I2, and I3?
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Apply KCL at the Top Node
I= I1 + I2 + I3
V
I1 
R1
V
I2 
R2
V
I3 
R3
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Solve for V
1
V V V
1
1
I 

 V  
 
R1 R2 R3
 R1 R2 R3 
1
V I
1
1
1


R1 R2 R3
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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
ECE201 Lect-3
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Current Divider
• This leads to a current divider equation for
three or more 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.
ECE201 Lect-3
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Example: More Than One Source
Is1
Is2
I1
R1
I2
R2
+
V
–
How do we find I1 or I2?
ECE201 Lect-3
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Apply KCL at the Top Node
I1 + I2 = Is1 - Is2
1
V V
1 
I s1  I s 2  
 V   
R1 R2
 R1 R2 
R1 R2
V  I s1  I s 2 )
R1  R2
ECE201 Lect-3
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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.
ECE201 Lect-3
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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
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Class Examples
• Learning Extension E2.10
• Learning Extension E2.11
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