Single Loop Circuits (2.3); Single-Node-Pair Circuits (2.4)

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Single Loop Circuits (2.3) Single-Node-Pair
Circuits (2.4)

• Prof. Phillips
• January 29, 2003

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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
I
228W
228W

50 Bulbs Total
120V
228W
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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
I



228W
228W
228I


120V

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
• This approach works for any single loop circuit
  with voltage sources and resistors.
• Resistors in series

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Voltage Division

• Consider two resistors in series with a voltage
  v(t) across them

R1
v1(t)

v(t)

R2
v2(t)
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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

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Class Examples
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Example 2 Light Bulbs in Parallel

• How do we find I1 and I2?

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Apply KCL at the Top Node
I I1 I2 Ohms Law
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Solve for V
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

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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

I1
I2
I3
I
R2
R1
R3
V

• How do we find I1, I2, and I3?

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Apply KCL at the Top Node
I I1 I2 I3
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Solve for V
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Req
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 equations similarity to the voltage
  divider equation.

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Example More Than One Source

I1
I2
Is2
Is1
R1
R2
V

• How do we find I1 or I2?

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Apply KCL at the Top Node
I1 I2 Is1 - Is2
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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 Ohms
  law.

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In General Current Division

• Consider N resistors in parallel
• Special Case (2 resistors in parallel)

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Class Examples