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
4
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
5
Common dry cell battery
PHY2049: Chapter 27
6
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
7
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
9
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
13
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
16
Resistance Example
ÎFind
net resistance of the circuit
connected to the battery. Each
resistance has R = 3 kΩ
PHY2049: Chapter 27
17
Resistance Example (1+2)
ÎCombine
‹ R12
#1 & #2 in series
= 6 kΩ
PHY2049: Chapter 27
18
Resistance Example (1+2+3)
ÎCombine
‹ R123
6kΩ & #3 in parallel
= 2 kΩ
PHY2049: Chapter 27
19
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
21
Resistance Example (1+2+3+4+5+6)
ÎCombine
‹ R123456
A
1.875kΩ & #6 in series
= 4.875 kΩ
C
PHY2049: Chapter 27
22
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
24
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
25
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
26
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
28
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
30