Circuits
PHY2049: Chapter 27
1
What You Already Know ÎNature
of current
ÎCurrent ÎDrift
density
speed and current
ÎOhm’s
law
ÎConductivity ÎCalculating ÎPower
and resistivity
resistance from resistivity
in electric circuits
PHY2049: Chapter 27
2
Chapter 27: Electric Circuits ÎWork,
energy and EMF
ÎSingle
loop circuits
ÎMultiloop
circuits
ÎAmmeters ÎRC
and voltmeters
circuits and time constant
PHY2049: Chapter 27
3
Reading Quiz for Chapter 27 ÎThe
electric current is defined as:
(1)
amount of charge per time (2) amount of charge per area (3) amount of charge per volume ÎWhen
resistors are connected in series
(1)
the current in each resistor is different (2) the current in each resistor is the same (3) the voltage in each resistor is the same ÎWhich
of the following is not related to Kirchhoff’s Rules?
(1)
conservation of charge (2) conservation of energy (3) conservation of momentum
PHY2049: Chapter 27
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EMF ÎEMF
device performs work on charge carriers
Converts
energy to electrical energy Moves carriers from low potential to high potential Maintains potential across terminals ÎVarious
types of EMF devices
Battery Generator Fuel
cell Solar cell Thermopile ÎExample:
Electrolytic reaction Magnetic field Oxidation of fuel Electromagnetic energy Nuclear decay
battery
Two
electrodes (different metals) Immersed in electrolyte (dilute acid) One electrode develops + charge, the other – charge PHY2049: Chapter 27
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Common dry cell battery
PHY2049: Chapter 27
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Electrons in the wire ÎIf
the electrons move so slowly through the wire, why does the light go on right away when we flip a switch? Household
wires have almost no resistance The electric field inside the wire travels much faster Light switches do not involve currents None of the above
PHY2049: Chapter 27
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Electrons in the wire, part 2 ÎOkay,
so the electric field in a wire travels quickly. But, didn’t we just learn that E = 0 inside a conductor? True,
it can’t be the electric field after all!! The electric field travels along the outside of the conductor E = 0 inside the conductor applies only to static charges None of the above
PHY2049: Chapter 27
8
Series Circuit ÎSimple
series situation
1
battery and two resistors R1 and R2 Common current I ÎTotal
voltage E = V1 + V2
= IR1 + IR2 ≡ IRs Rs = R1 + R2 E
ÎSo
equivalent resistance is the sum
Works
for any number of resistors Rs = R1 + R2 + R3 + R4 + …
PHY2049: Chapter 27
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Resistors in series ÎEMF
of battery is 12 V, 3 identical resistors. What is the potential difference across each resistor? 12
V 0 V 3 V 4 V
Current is the same, so voltage across each is the same
R
PHY2049: Chapter 27
R
R
10
Resistors in series ÎIf
the light bulbs are all the same in each of these two circuits, which circuit has the higher current? circuit
A circuit B both the same ÎIn
which case is each light bulb brighter?
circuit
A circuit B both the same
A
B PHY2049: Chapter 27
11
Real EMF Sources: Internal Resistance batteries have small internal resistance
Lowers
i=
effective potential delivered to circuit
E r+R
Vb
r
Veff = Vb − Va = E − ir
Va
R
E =E− r r+R
C
Veff
C
ÎReal
E = 1+ r / R
This is the voltage measured across the terminals! PHY2049: Chapter 27
12
Internal Resistance Example ÎLoss
ÎE
of voltage is highly dependent on load E 12 Veff = = 1 + r / R 1 + 0.1/ R
= 12V, r = 0.1Ω, R = 100Ω
Loss
ÎE ÎE
Veff = 12 /1.1 = 10.9V
of 1.1V
= 12V, r = 0.1Ω, R = 0.5Ω
Loss
Veff = 12 /1.01 = 11.9V
of 0.1V
= 12V, r = 0.1Ω, R = 1Ω
Loss
ÎE
of 0.01V
= 12V, r = 0.1Ω, R = 10Ω
Loss
Veff = 12 /1.001 = 11.99V
Veff = 12 /1.2 = 10.01V
of 2.0V PHY2049: Chapter 27
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Heating From Internal Resistance ÎHeating
of EMF source: P = i2r
Heating
ÎE
is extremely dependent on load
= 12V, r = 0.1Ω
R
= R = R = R =
100Ω 10Ω 1.0Ω 0.5Ω
Vba Vba Vba Vba
= = = =
11.99V 11.9V 10.9V 10.0V
I I I I
= = = =
0.12 A 1.19 A 10.9 A 20 A
P P P P
= = = =
0.0014 W 0.14 W 11.9 W 40 W
2
14.4 ⎛ E ⎞ P=i r =⎜ ⎟ r = 2 r R + ⎝ ⎠ ( R + 0.1) 2
PHY2049: Chapter 27
14
Resistors in Parallel Î Current
splits into several branches. Total current is conserved
I = I1 + I2
Î Potential
resistor
difference is same across each
V = V1 = V2
a
I
V d
V V V = + R p R1 R2
a
Rp = equivalent resistance. Works for any number of resistors PHY2049: Chapter 27
I2
R1
R2
I I I
V
1 1 1 = + R p R1 R2
I1
Rp d
I
15
Resistors in Parallel ÎAs
more resistors R are added in parallel to the circuit, how does total resistance between points P and Q change? (a)
increases (b) remains the same (c) decreases
ÎIf
the voltage between P & Q is held constant, and more resistors are added, what happens to the current through each resistor? (a)
increases (b) remains the same (c) decreases
Overall current increases, but current through each branch is still V/R. PHY2049: Chapter 27
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Resistance Example ÎFind
net resistance of the circuit connected to the battery. Each resistance has R = 3 kΩ
PHY2049: Chapter 27
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Resistance Example (1+2) ÎCombine R12
#1 & #2 in series
= 6 kΩ
PHY2049: Chapter 27
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Resistance Example (1+2+3) ÎCombine R123
6kΩ & #3 in parallel
= 2 kΩ
PHY2049: Chapter 27
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Resistance Example (1+2+3+4) ÎCombine R1234
2kΩ & #4 in series
= 5 kΩ
B
A
A
C
PHY2049: Chapter 27
C
20
Resistance Example (1+2+3+4+5) ÎCombine R12345
5kΩ & #5 in parallel
= 1.875 kΩ
A A
C
C
PHY2049: Chapter 27
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Resistance Example (1+2+3+4+5+6) ÎCombine R123456
A
1.875kΩ & #6 in series
= 4.875 kΩ
C
PHY2049: Chapter 27
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Circuits ÎIf
the light bulbs are all the same in each of these two circuits, which circuit has the higher current? (a)
circuit A (b) circuit B (c) both the same B draws twice the current as A ÎIn
which case is each light bulb brighter?
A
(a)
circuit A (b) circuit B (c) both the same Current through each branch is unchanged (V/R)
B PHY2049: Chapter 27
23
Light Bulb Problem ÎTwo
light bulbs operate at 120 V, one with a power rating of 25W and the other with a power rating of 100W. Which one has the greater resistance? (a)
the one with 25 W (b) the one with 100 W (c) both have the same resistance
ÎWhich
carries the greater current?
(a)
the one with 25 W (b) the one with 100 W (c) both have the same current
P = I2R = V2/R, so 100W bulb has ¼ the resistance of the 25W bulb and carries 4x the current.
PHY2049: Chapter 27
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Dimmer ÎWhen
you rotate the knob of a light dimmer, what is being changed in the electric circuit? (a) (b) (c) (d) (e)
the voltage the resistance the current both (a) and (b) both (b) and (c)
House voltage is always 110 – 120V. Turning the knob increases the circuit resistance and thus lowers the current.
PHY2049: Chapter 27
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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
PHY2049: Chapter 27
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Household Circuits Î All
devices are added in parallel.
Î Overload:
too many devices that require a lot of current can draw more current than wires can handle. Overheating of wires Fire hazard!
PHY2049: Chapter 27
27
Resistors ÎCurrent
flows through a light bulb. If a wire is now connected across the bulb as shown, what happens? (a)
All the current continues to flow through the bulb (b) Current splits 50-50 into wire and bulb (c) Almost all the current flows through the wire (d) None of the above Current is maintained by external voltage source. The wire “shunt” has almost no resistance and it is in parallel with a bulb having resistance. Therefore all the current follows the zero (or extremely low) resistance path.
PHY2049: Chapter 27
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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).
PHY2049: Chapter 27
29
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 PHY2049: Chapter 27
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