Transcript W10D2_DC_Circuits_v02_no answers_jwb

Faraday’s Law of Induction
If C is a stationary closed curve and S is a
surface spanning C then
d
E
×
d
s
=
òC
dt
B
×
d
A
òò
S
The changing magnetic flux through S
induces a non-electrostatic electric field
whose line integral around C is non-zero
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Problem: Calculating Induced
Electric Field
d
E×ds = B×dA
ò
òò
dt open surface
closed path
Consider a uniform magnetic field
which points into the page and is
confined to a circular region with
radius R. Suppose the magnitude
increases with time, i.e. dB/dt > 0.
Find the magnitude and direction of
the induced electric field in the
regions (i) r < R, and (ii) r > R. (iii)
Plot the magnitude of the electric
field as a function r.
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W10D2
DC Circuits
Today’s Reading Assignment W10D2 DC Circuits &
Kirchhoff’s Loop Rules Course Notes: Sections 7.1-7.5
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Announcements
PS 8 due Week 11 Tuesday at 9 pm in boxes outside 32-082 or
26-152
Next Reading Assignment W10D3 PS07: PhET: Building a
Circuit 7.1-7.5, 7.10
Exam 3 Thursday April 18 7:30 pm –9:30 pm
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Outline
DC Circuits
Kirchoff’s Laws
Electrical Power
Measuring Voltage and Current
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DC Circuits
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Examples of Circuits
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Symbols for Circuit Elements
Battery
Resistor
Capacitor
Switch
Equipotential
Junction
Branch
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Electromotive Force (EMF)
The work per unit charge around a closed path done
is called electromotive force EMF. This is a bad
name because it is not a force. Let f s denote the
force per unit charge, then the EMF is
e= ò
fs × d s
closd path
If a conducting closed path is present then
e = IR
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Ideal Battery
1) Inside Battery: chemical force fs non-zero inside battery
(zero outside) moves charges through a region in which
static electric field Estatic opposes motion Estatic = -fs
2) Ideal battery: net force on charges is zero
3) Potential difference between its terminals
+
+
-
-
V (+) -V (-) = - ò Estatic × d s = ò fs × d s
4) Extend path through
external circuit where fs = 0
+
ò f ×ds = ò f ×ds = e
s
s
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Sign Conventions - Battery
Moving from the negative to positive terminal of a
battery increases your potential
DV = Vb - Va
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Sign Conventions - Resistor
Moving across a resistor in the direction of current
decreases your potential
DV = Vb - Va
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Internal Resistance
Real batteries have an internal resistance, r, which is
small but non-zero
Terminal voltage:
DV = Vb - Va = e - I r
(Even if you short the leads you don’t get infinite current)
Series vs. Parallel
Series
Parallel
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Resistors In Series
The same current I must flow through both resistors
DV = I R1 + I R2 = I(R1 + R2 ) = I Req
Req = R1 + R2
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Resistors In Parallel
Voltage drop across the resistors must be the same
DV = DV1 = DV2 = I1 R1 = I 2 R2 = IReq
DV DV DV
+
=
I = I1 + I 2 =
R1
R2
Req
1
1
1
=
+
Req R1 R2
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Measuring Voltage & Current
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Measuring Potential Difference
A voltmeter must be hooked in parallel across the
element you want to measure the potential difference
across
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Reffective
1
1
= +
R RVoltmeter
1
R
Voltmeters have a very large resistance, so that
they don’t affect the circuit too much
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Measuring Current
An ammeter must be hooked in series with the
element you want to measure the current through
Reffective = R + Rammeter
R
Ammeters have a very low resistance, so that
they don’t affect the circuit too much
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Measuring Resistance
An ohmmeter must be hooked in parallel across the
element you want to measure the resistance of
Here we are measuring R1
Ohmmeters apply a voltage and measure the
current that flows. They typically won’t work if the
resistor is powered (connected to a battery)
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Concept Question: Bulbs &
Batteries
An ideal battery is hooked to a light
bulb with wires. A second identical
light bulb is connected in parallel to
the first light bulb. After the second
light bulb is connected, the current
from the battery compared to when
only one bulb was connected.
1.
2.
3.
4.
Is Higher
Is Lower
Is The Same
Don’t know
P18- 21
Concept Question: Bulbs &
Batteries
An ideal battery is hooked to a
light bulb with wires. A second
identical light bulb is connected in
series with the first light bulb. After
the second light bulb is connected,
the current from the battery
compared to when only one bulb
was connected.
1. Is Higher
2. Is Lower
3. Is The Same
P18- 22
Electrical Power
Power is change in energy per unit time
So power to move current through circuit elements:
(
)
d
d
dq
P= U =
qDV =
DV
dt
dt
dt
P  I V
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Power - Battery
Moving from the negative to positive terminal of a
battery increases your potential. If current flows
in that direction the battery supplies power
I
Psupplied  I V  I 
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Power – Resistor (Joule Heating)
Moving across a resistor in the direction of current
decreases your potential. Resistors always
dissipate power
V
Pdissipated  I V  I R 
R
2
2
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Concept Question: Power
An ideal battery is hooked to a light bulb
with wires. A second identical light bulb
is connected in parallel to the first light
bulb. After the second light bulb is
connected, the power output from the
battery (compared to when only one
bulb was connected)
1.
2.
3.
4.
5.
Is four times higher
Is twice as high
Is the same
Is half as much
Is ¼ as much
P18- 26
Concept Question: Power
An ideal battery is hooked to a light
bulb with wires. A second identical
light bulb is connected in series with
the first light bulb. After the second
light bulb is connected, the light
(power) from the first bulb (compared
to when only one bulb was connected)
1.
2.
3.
4.
5.
Is four times higher
Is twice as high
Is the same
Is half as much
Is ¼ as much
P18- 27
Kirchhoff’s Loop Rules
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Current Conservation
Sum of currents entering any junction in a circuit
must equal sum of currents leaving that junction.
I1 = I 2 + I3
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Concept Question: Measuring
Current
If R1 > R2, compare the
currents measured by the
three ammeters:
1.
2.
3.
4.
5.
6.
7.
A1 > A 2 > A 3
A2 > A 1 > A 3
A3 > A 1 > A 2
A3 > A 2 > A 1
A3 > A 1 = A 2
None of the above
Not enough information is given.
P18- 30
Sum of Potential Differences
Around a Closed Path
dB
NO TIME DEPENDENCE!
0
dt
 Vi    Estatic  d s  0
i
Closed
Path
Sum of potential differences across all elements
around any closed circuit loop must be zero.
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Sum of Potential Differences
Around a Closed Path
 Vi    Estatic  d s  0
i
Closed
Path
Sum of potential differences across all elements
around any closed circuit loop must be zero.
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Demonstration:
Five Different Types of Resistance
F13
http://tsgphysics.mit.edu/front/?page=demo.php&letnum=F 13&show=0
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Steps of Solving Circuit Problem
1.
2.
3.
4.
5.
Straighten out circuit (make squares)
Simplify resistors in series/parallel
Assign current loops (arbitrary)
Write loop equations (1 per loop)
Solve
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Worked Example: Circuit
What is current through each branch of the circuit
shown in the figure below?
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Worked Example: Simple Circuit
Current conservation at a:
I1 = I 2 + I3
Lower Loop: Start at a and
circulate clockwise. Then
e - I 2 2R1 + I 3 3R1 = 0 Þ
I 3 = (2 / 3)I 2 - (e / 3R1 )
Upper Loop: Start at a and circulate clockwise. Then
2e - I1 R1 - I 2 2R1 = 0 Þ I1 = (2e / R1 ) - 2I 2
I1 = I 2 + I3 Þ (2e / R1 ) - I 2 = I 2 + (2 / 3)I 2 - (e / 3R1 )
Solve for I2:
I 2 = 7e / 11R1 Þ I 3 = e / 11R1 and I 3 = 8e / 11R1
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Group Problem: Four Resistors
Four resistors are
connected to a battery
as shown in the figure. The
current in the battery is I,
the battery emf is ε, and
the resistor values are R1 =
R, R2 = 2R, R3 = 4R, R4 =
3R.
Determine the current in
each resistor in terms of I.
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Demonstration:
Wheatstone Bridge F14
http://tsgphysics.mit.edu/front/?page=demo.php&letnum=F 14&show=0
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Group Problem: Wheatstone Bridge
A circuit consisting of two
resistors with R1=6.0 ohms and
R2=1.5 ohms, a variable resistor,
with resistance Rvar, a resistor of
unknown value Ru, and 9.0 volt
battery, are connected as shown
in the figure. When Rvar is
adjusted to 12 ohms, there is
zero current through the
ammeter. What is the unknown
resistance Ru?
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