Single Loop Circuits (2.3); Single-Node-Pair Circuits (2.4)
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Transcript of Single Loop Circuits (2.3); Single-Node-Pair Circuits (2.4)
ECE201 Lect-3 1
Single Loop Circuits (2.3); Single-Node-Pair Circuits (2.4)
Prof. Phillips
January 29, 2003
ECE201 Lect-3 2
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.
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Example: Christmas Lights
+–120V
228
50 Bulbs Total
228
228
I
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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.
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228I + 228I + … + 228I -120V = 0
I = 120V/(50 228) = 10.5mA
+–120V
228
228
228
I + –
228I
+
–
228I
228I
+
–
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Some Comments
• We can solve for the voltage across each light bulb:
V = IR = (10.5mA)(228 = 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:
• This approach works for any single loop circuit with voltage sources and resistors.
• Resistors in series
sresistanceofsumsourcesvoltageofsum
RV
tij
Si )(
jNseries RRRRR 21
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Voltage Division
Consider two resistors in series with a voltage v(t) across them:
R1
R2
–
v1(t)
+
+
–
v2(t)
+
–
v(t)21
11 )()(
RR
Rtvtv
21
22 )()(
RR
Rtvtv
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In General: Voltage Division
Consider N resistors in series:
Source voltage(s) are divided between the resistors in direct proportion to their resistances
j
iSR R
RtVtV
ki)()(
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Class Examples
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Example: 2 Light Bulbs in Parallel
How do we find I1 and I2?
I R1 R2 V
+
–
I1 I2
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Apply KCL at the Top Node
I= I1 + I2
Ohm’s Law:1
1 R
VI
22 R
VI
I R1 R2 V
+
–
I1 I2
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Solve for V
212121
11
RRV
R
V
R
VIII
21
21
21
111
RR
RRI
RR
IV
Rearrange
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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:
Definition: Parallel - the elements share the same two end nodes
21
21
RR
RRReq
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21
2
1
21
21
11 RR
RI
R
RRRR
I
R
VI
Now to find I1
• 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
How do we find I1, I2, and I3?
I R2 V
+
–
R1
I1 I2
R3
I3
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Apply KCL at the Top Node
I= I1 + I2 + I3
11 R
VI
22 R
VI
33 R
VI
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Solve for V
321321
111
RRRV
R
V
R
V
R
VI
321
1111
RRR
IV
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Req
321
1111
RRR
Req
iMpar RRRRR
11111
21
Which is the familiar equation for parallel resistors:
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Current Divider
• This leads to a current divider equation for three or more parallel resistors.
• For 2 parallel resistors, it reduces to a simple form.
• Note this equation’s similarity to the voltage divider equation.
j
parSR R
RII
j
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Is2 VR1 R2
+
–
I1 I2
Example: More Than One Source
How do we find I1 or I2?
Is1
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Apply KCL at the Top Node
I1 + I2 = Is1 - Is2
212121
11
RRV
R
V
R
VII ss
21
2121 RR
RRIIV ss
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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.
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In General: Current Division
Consider N resistors in parallel:
Special Case (2 resistors in parallel)
iNpar
j
parSR
RRRRR
R
Rtiti
kj
11111
)()(
21
21
2)()(1 RR
Rtiti SR
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Class Examples