First, we will consider circuits with batteries and resistors.
Quick Overview of Electric Circuit Components.
Capacitor – stores charge and potential
energy, measured in Farads (F)
V=5V
Battery – generates a constant electrical
potential difference (ΔV) across it.
Measured in Volts (V).
I
R
Resistor – resists flow of charge due to
scattering; dissipates energy.
Measured in Ohms (Ω)
1
What happens to the resistor (light bulb) when we turn it on?
2
The moving charges (for example electrons) are shifting down
in potential energy.
ΔPE = ΔU = qΔV
It emits light. It also gets hot ! Heat is a form of energy.
ΔN q
ΔQ
=q
Δt
Δt
ΔU qΔN q ΔV ⎛ ΔN q ⎞
⎟⎟ΔV = iV
P=
=
= ⎜⎜ q
Δt
Δt
⎝ Δt ⎠
Power P = Energy/Time = [Joules/second] = [Watts]
i=
Inside a current carrying resistor, electrostatic potential
energy is converted into thermal energy (heat).
3
Power
Where did the potential energy go?
i
P = iV
It cannot increase the drift velocity since that is set by the
electric field magnitude. Does this mean Kinetic Energy is
constant?
V or ΔV
Energy goes into heat. Heat is in fact random (thermal)
motion of particles, which is microscopic kinetic energy.
However, these velocities are random and do not change
the overall drift velocity.
If the resistor is Ohmic, then V=iR and thus
P = iV = i 2 R =
4
2
V
R
5
6
1
Clicker Question
Demonstrations of heat generation in resistors
“100 Watt” light bulb
For all household appliances, they have:
ΔV = 120V
What is the approximate resistance of the filament (at
operating temperature)?
A) R = 100 Ohms
B) R = 144 Ohms
In fact, 95% is heat, and
C)R = 1250 Ohms
only 5% of energy is in
D)None of the Above
the light.
7
Clicker Question
Combinations of Resistors:
R1
8
i1
R2
i2
i3
R1
R2
R3
Take the case where R1 = 5 Ω, R2 = 10 Ω, and R3 = 20 Ω.
If the current i1 = 5.0 Amps, what are the other currents?
A) i2 = 10 Amps, i3 = 20 Amps
B) i2 = 2.5 Amps, i3 = 1.25 Amps
C) i2 = 1.25 Amps, i3 = 0.30 Amps
D) i2 = 0.0 Amps, i3 = 0.0 Amps
i1 = i2 = i3 = 5.0 Amps
E) None of the above answers.
R3
Resistors in Series
9
10
Resistors in Series
Calculate i and V across each resistor and then P.
R1
R2
R3
Ibat
Req = R1 + R2 + R3 = ∑ Ri
RA
IA
RA = 1.0 Ω
IB
RB = 3.0 Ω
V
V = 6 Volts
i
Adding resistors in series always increases the total resistance.
11
RB
12
2
Clicker Question
What is the equivalent resistance?
Two different wires are connected to a battery in
series (in a chain, one after another). How does
the current in upper wire A compare to the current
in lower wire B?
V
A) iA > iB
B) iA < iB
C) iA = iB
D) Impossible to determine.
Req = R1 + R2 = 1.0Ω + 3.0Ω = 4.0Ω
Ibat
RA
IA
RB
IB
Ibat
What is the current through the circuit?
Assuming everything is Ohmic, V=iR
i = i A = iB =
RA
IA
RB
IB
V
V
6V
=
= 1.5 Amps
Req 4.0Ω
13
14
What is the Power lost to heat in each resistor?
Ibat
V2
P = iV = i R =
R
RA
IA
RB
IB
Demonstration with two wires of different resistance.
V
2
Which section of wire is going to generate the most heat?
V
We already know current (i) and resistance (R).
Ibat
2
RB
PB = iB RB = (1.5 A) 2 (3.0Ω) = 6.75Watts
RA
PA = i A RA = (1.5 A) 2 (1.0Ω) = 2.25Watts
2
IA
IB
15
Clicker Question
What about the Voltage change across each resistor?
Ibat
RA
IA
RA = 1.0 Ω, iA=1.5 Amps
IB
RB = 3.0 Ω, iB = 1.5 Amps
V
V = 6 Volts
RB
16
VA = ΔVA = i A RA = (1.5 A)(1.0Ω) = 1.5Volts
VB = ΔVB = iB RB = (1.5 A)(3.0Ω) = 4.5Volts
17
If we consider the Voltage change starting
Ibat
at Point P and going all the way around the
circuit loop back to Point P, what is the total V
ΔV?
Α)ΔV = +6.0 Volts
Β) ΔV = +1.5 Volts
C) ΔV = -4.5 Volts
D) ΔV = 0.0 Volts
Point P
E) ΔV = -6.0 Volts
IA
RA
IB
RB
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3
Voltage
P2
Kirchhoff’s Voltage Loop Law:
P3
Ibat
IA
RA
6V
The change in Voltage around any closed loop must be zero.
4.5V
V
P4
IB
RB
P1
P5
3.0V
Kirchhoff’s Current Junction Law:
1.5V
0.0V
1
2
3
4
1 Position
In Loop
5
Define V=0 at P1.
In steady state, the current going into a junction (or point) must
equal the current going out of that junction (or point).
19
Clicker Question
20
Clicker Question
We start with the left circuit with one lightbulb (A). The
brightness of the bulb directly reflects the power. If we add a
second bulb (B) as shown on the right, what happens to the
bulbs?
A) Bulb A is equally
bright.
B) Bulb A is dimmer
V
V
than before
A
A
B
C)Bulb A is brighter
than before
Two light bulbs, A and B, are in series, so they carry the
same current. Light bulb A is brighter than B.
Which bulb has higher resistance?
A) A
B) B
C) Same resistance.
Answer: Bulb A has higher resistance.
Since the resistors are in series, they
have the same current I. According to
P = I2 R, if I = constant, then higher R
gives higher P.
V
A
B
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22
Resistors in Parallel
Demonstration with two light bulbs in series.
R1
R2
V
A
B
R3
23
24
4
Clicker Question
R1
R2
R3
V
If we connect a battery with
V = 6 Volts with three resistors
R1 = 10 Ω, R2 = 20 Ω, R3 = 30 Ω,
which of the following is true?
A) The current through each resistor
is the same.
B) The Voltage change through
each resistor is the same.
C)The resistance of each resistor is
the same.
D)None of the above are true.
What if we apply the current junction rule at this point?
i1
R1
i2
R3
i
i = i1 + i2 + i3
R2
i3
V
If all three resistors were the
same then:
1
i1 = i2 = i3 = i
3
25
Equivalent of resistors in parallel
26
Resistors in Series
Resistors in Parallel
Req = R1 + R2 + R3
Req =
R1
R2
R3
Req =
1
1
1
1
+
+
R1 R2 R3
or
1
1
1
1
= +
+
Req R1 R2 R3
1
1
1
1
+
+
R1 R2 R3
Always increases the resistance!
Always decreases the resistance!
27
Which is the best way to wire a house?
28
Traffic Analog !
Resistors in
parallel are like
having more lanes
on the highway.
This reduces the
resistance for
getting from one
place to another.
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30
5
Clicker Question
Clicker Question
We start with the left circuit with one lightbulb (A). The
brightness of the bulb directly reflects the power. If we add a
second bulb (B) as shown on the right, what happens to the
bulbs?
B
A) Bulb A is equally
bright.
A
B) Bulb A is dimmer
A
than before
C)Bulb A is brighter
V=12V
V=12V
than before
What is the electric potential difference across the upper light
bulb (resistor)? Think about our Voltage Loop Rule.
A) |V| = 0 Volts
B) |V| = 6 Volts
C)|V| = 12 Volts
D)|V| = 24 Volts
E) None of the above answers.
V=12V
31
Clicker Question
If each of these six light bulbs is
identical, which bulb is going to
be the brightest?
A) Bulb A
B) Bulb B
C)Bulb C
D)Bulb D
E) Bulb E
32
Clicker Question
The three light bulbs A, B, and C are identical. How does the
brightness of bulbs B and C together compare with the
brightness of bulb A?
A) Total power in B+C = power in A.
A
B) Total power in B+C > power in A.
C) Total power in B+C < power in A.
B
C
V=12V
Answer: Use P = V2/Rtot For bulbs B
and C, Rtot = 2R.
Total power in B+C < power in A.
33
Clicker Question
In the circuit below, what happens to the brightness of bulb 1,
when bulb 2 burns out? (When a bulb burns out, its resistance
becomes infinite.)
2
A) Bulb 1 gets brighter
B) Bulb 1 gets dimmer.
C) It's brightness remains the same.
3
1
(Hint: What happens to the current from
the battery when bulb 2 burns out.)
34
Answers: Bulb 1 gets dimmer! When bulb 2 burns outs, the
filament inside breaks and R2 becomes infinitely large. The
total equivalent resistance which the battery sees increases
(since bulb 2 is gone, there are fewer paths for the current
flow, so less flow, more total resistance.) Since the battery
sees a larger R-tot, the current from the battery I-tot = V/Rtot
is reduced. Less current from the battery means less current
2
through bulb 1, less light.
3
1
V=12V
V=12V
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