Circuits
ÎThe
light bulbs in the circuits below are identical. Which
configuration produces more light?
‹ (a)
circuit I
‹ (b) circuit II
‹ (c) both the same
Circuit II has ½ current of each branch
of circuit I, so each bulb is ¼ as bright.
The total light in circuit I is thus 4x that
of circuit II.
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Circuits
ÎThe
three light bulbs in the circuit are identical. The
current flowing through bulb B, compared to the current
flowing through bulb A, is
‹ a)
‹ b)
‹ c)
‹ d)
‹ e)
4 times as much
twice as much
the same
half as much
1/4 as much
Branch of circuit A has ½ resistance,
thus it has 2x current.
PHY2054: Chapter 18
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Circuits
ÎThe
three light bulbs in the circuit are identical. What is
the brightness of bulb B compared to bulb A?
‹ a)
‹ b)
‹ c)
‹ d)
‹ e)
4 times as much
twice as much
the same
half as much
1/4 as much
Use P = I2R. Thus 2x current in A
means it is 4x brighter.
PHY2054: Chapter 18
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More Complicated Circuits
ÎParallel
ÎUse
and series rules are not enough!
Kirchoff’s rules
PHY2054: Chapter 18
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Problem Solving Using Kirchhoff’s Rules
ÎLabel
the current in each branch of the circuit
‹ Choice
of direction is arbitrary (signs work out in the end)
‹ Apply junction rule at each junction: I = I1 + I2
I
ÎApply
loop rule to each loop (follow in one direction)
I1
I2
‹ Resistors:
if loop direction in current direction, voltage drop
‹ Batteries: if loop direction in “normal” direction, voltage gain
‹ Sum of all voltages = 0 around loop
+V - IR
ÎSolve
equations simultaneously
‹ You
ÎSee
need as many equations as you have unknowns
example next slide
PHY2054: Chapter 18
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Example 1: Applying Kirchhoff’s Rules
ÎDetermine
magnitudes and directions of all currents
‹ Take
two loops, 1 and 2, as shown
‹ Define currents: I1, I2, I3
Use I1 = I2 + I3
9V
22 Ω
I2
2
I3
15 Ω
1
+6 − 15I 3 = 0
2
−22 I 2 + 9 + 15 I 3 = 0
I1
6V
I 2 = 15 / 22 = 0.68
1
I 3 = 6 /15 = 0.40
I1 = I 2 + I 3 = 1.08
I1
PHY2054: Chapter 18
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Example 2: Applying Kirchhoff’s Rules
Use I3 = I1 + I2
1
+10 − 6 I1 + 14 + 4 I 2 = 0
2
+10 − 6 I1 − 2 I 3 = 0
I1 = 2
I 2 = −3
I3 = I1 + I 2 = −1
PHY2054: Chapter 18
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Circuits
ÎWhich
of the equations is valid for the circuit loop below?
2 − I1 − 2I2 = 0
‹ b) 2 − 2I1 − 2I2 − 4I3 = 0
‹ c) 2 − I1 − 4 − 2I2 = 0
‹ d) I3 − 2I2 − 4I3 = 0
‹ e) 2 − 2I1 − 2I2 − 4I3 = 0
‹ a)
PHY2054: Chapter 18
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Circuit Problem (1)
ÎThe
light bulbs in the circuit are identical. What happens
when the switch is closed?
‹ a)
both bulbs go out
‹ b) the intensity of both bulbs increases
‹ c) the intensity of both bulbs decreases
‹ d) nothing changes
(a)
(b)
Before: Potential at (a) is 12V, but so
is potential at (b) because equal
resistance divides 24V in half. When the
switch is closed, nothing will change
since (a) and (b) are still at same potential.
PHY2054: Chapter 18
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Circuit Problem (2)
ÎThe
light bulbs in the circuit shown below are identical.
When the switch is closed, what happens to the intensity
of the light bulbs?
‹ a)
bulb A increases
‹ b) bulb A decreases
‹ c) bulb B increases
‹ d) bulb B decreases
‹ e) nothing changes
(a)
(b)
Before: Potential at (a) is 12V, but so
is potential at (b) because equal
resistance divides 24V in half. When
the switch is closed, nothing will
change since (a) and (b) are still at
same potential.
PHY2054: Chapter 18
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Circuit Problem (3)
ÎThe
bulbs A and B have the same R. What happens when
the switch is closed?
‹ a)
nothing happens
‹ b) A gets brighter, B dimmer
‹ c) B gets brighter, A dimmer
‹ d) both go out
Before: Bulb A and bulb B both have
18V across them.
(a)
(b)
24V
After: Bulb A has 12V across it and
bulb B has 24V across it (these
potentials are forced by the batteries).
PHY2054: Chapter 18
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Wheatstone Bridge
ÎAn
ammeter A is connected between points a and b in the
circuit below, in which the four resistors are identical.
What is the current through the ammeter?
‹ a)
I/2
‹ b) I / 4
‹ c) zero
‹ d) need more information
The parallel branches have the same
resistance, so equal currents flow in
each branch. Thus (a) and (b) are at
the same potential and there is no
current flow across the ammeter.
PHY2054: Chapter 18
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