Power lines
ÎAt
large distances, the resistance of power lines becomes
significant. To transmit maximum power, is it better to
transmit high V, low I or high I, low V?
‹ (a)
high V, low I
‹ (b) low V, high I
‹ (c) makes no difference
ÎWhy
Power loss is I2R, so want
to minimize current.
do birds sitting on a high-voltage power line survive?
‹ They
are not touching high and low potential simultaneously to
form a circuit that can conduct current
PHY2054: Chapter 18
18
Resistors
ÎCurrent
flows through a light bulb. If a wire is now
connected across the bulb as shown, what happens?
‹ (a)
Bulb remains at same brightness
‹ (b) Bulb dims to 1/4 its former brightness (1/2 current)
‹ (c) Bulb goes out
‹ (d) None of the above
The wire “shunt” has almost no resistance and
it is in parallel with a bulb having resistance.
Therefore voltage across shunt (and bulb ) is ~ 0.
Thus almost all the current follows the zero
(or extremely low) resistance path.
PHY2054: Chapter 18
19
Circuits
ÎTwo
light bulbs A and B are connected in series to a
constant voltage source. When a wire is connected across
B, what will happen to bulb A?
‹ (a)
burns more brightly than before
‹ (b) burns as brightly as before
‹ (c) burns more dimly than before
‹ (d) goes out
The wire shunt effectively eliminates the
second resistance, hence increasing the
current in the circuit by 2x. The first
bulb burns 4x brighter (I2 R).
PHY2054: Chapter 18
20
Voltage Drops in Series Circuit
Î Series
circuit
‹ Battery
Î Current
V, resistors R1 and R2
i = V / (R1 + R2)
‹ Voltage
across R1 = IR1 = V * R1 / (R1 + R2)
‹ Voltage across R2 = IR2 = V * R2 / (R1 + R2)
Î So
voltage drops in proportion to resistance!
‹ Let
V = 24, R1 = 8Ω, R2 = 4Ω
‹ V1 = 24 * 8 / 12 = 16
‹ V1 = 24 * 4 / 12 = 8
Î Useful
when considering how resistors divide up voltage
PHY2054: Chapter 18
21
Circuits
ÎConsider
the network of resistors shown below. When the
switch S is closed, then:
‹ What
happens to the voltage across R1, R2, R3, R4?
‹ What happens to the current through R1, R2, R3, R4?
‹ What happens to the total power output of the battery?
‹ Let R1 = R2 = R3 = R4 = 90 Ω and V = 45 V. Find the current
through each resistor before and after closing the switch.
Before
¾I1 = 45/135 = 1/3
¾I2 = 0
¾I3 = I4 = 15/90=1/6
After
¾I1 = 45/120 = 3/8
¾I2 = I3 = I4 = 1/8
PHY2054: Chapter 18
22
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.
PHY2054: Chapter 18
23
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
24
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
25
More Complicated Circuits
ÎParallel
ÎUse
and series rules are not enough!
Kirchoff’s rules
PHY2054: Chapter 18
26
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
I
+V - IR
ÎSolve
equations simultaneously
‹ You
ÎSee
need as many equations as you have unknowns
example next slide
PHY2054: Chapter 18
27
Example 1: Applying Kirchhoff’s Rules
ÎDetermine
magnitudes and directions of all currents
‹ Take
two loops, 1 and 2, as shown
‹ Choose currents & directions!: 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
28
Example 2: Applying Kirchhoff’s Rules
¾ Pick currents as shown
¾ 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
29
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
30
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
31
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
32
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
33