10. resistor circuits. kirchhoff's laws
TRANSCRIPT
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University of Rhode Island University of Rhode Island
DigitalCommons@URI DigitalCommons@URI
PHY 204: Elementary Physics II -- Slides PHY 204: Elementary Physics II (2021)
2020
10. Resistor circuits. Kirchhoff's laws 10. Resistor circuits. Kirchhoff's laws
Gerhard Müller University of Rhode Island, [email protected]
Robert Coyne University of Rhode Island, [email protected]
Follow this and additional works at: https://digitalcommons.uri.edu/phy204-slides
Creative Commons License
This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 4.0
License.
Recommended Citation Recommended Citation Müller, Gerhard and Coyne, Robert, "10. Resistor circuits. Kirchhoff's laws" (2020). PHY 204: Elementary Physics II -- Slides. Paper 35. https://digitalcommons.uri.edu/phy204-slides/35https://digitalcommons.uri.edu/phy204-slides/35
This Course Material is brought to you for free and open access by the PHY 204: Elementary Physics II (2021) at DigitalCommons@URI. It has been accepted for inclusion in PHY 204: Elementary Physics II -- Slides by an authorized administrator of DigitalCommons@URI. For more information, please contact [email protected].
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Direct Current Circuit
Consider a wire with resistance R = ρ`/A connected to a battery.
• Resistor rule: In the direction of I across a resistor with resistance R, the electric potential drops:∆V = −IR.
• EMF rule: From the (−) terminal to the (+) terminal in an ideal source of emf, the potential rises: ∆V = E .• Loop rule: The algebraic sum of the changes in potential encountered in a complete traversal of any loop
in a circuit must be zero: ∑ ∆Vi = 0.
+ −emf
I
+ −emf
I
I
R
a b
ε
a b a
Va
Vb
−IR ε+ −
I
a b
circuit diagramphysical system electric potential
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Battery with Internal Resistance
• Real batteries have an internal resistance r.• The terminal voltage Vba ≡ Va −Vb is smaller than the emf E written on the label if a current flows
through the battery.• Usage of the battery increases its internal resistance.
• Current from loop rule: E − Ir− IR = 0 ⇒ I =E
R + r
• Current from terminal voltage: Vba = E − Ir = IR ⇒ I =VbaR
+ −emf
I
+ −emf
IR
ba
εI
r+ −
I
a b
electric potentialphysical system circuit diagram
a b
−Ir
a
Va
Vb
ε
−IR
.
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Resistor Circuit (4)
Consider the resistor circuit shown.
(a) Find the direction of the positive current (cw/ccw).(b) Find the magnitude of the current.(c) Find the voltage Vab = Vb −Va.(d) Find the voltage Vcd = Vd −Vc.
1Ω
12V
1Ω
2Ω
a
cd
b
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Resistor Circuit (6)
Consider the resistor circuit shown.
(a) Choose a current direction and use the loop rule to determine the current.(b) Name the direction of positive current (cw/ccw).(c) Find Vab ≡ Vb −Va along two different paths.
b
a
1Ω
6V
1Ω 18V
2Ω
24V
2Ω12V
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Power in Resistor Circuit
Battery in use• Terminal voltage: Vab = E − Ir = IR
• Power output of battery: P = VabI = E I− I2r
• Power generated in battery: E I• Power dissipated in battery: I2r
• Power transferred to load: P = I2R
b
a
ε
rR
Battery being charged:
• Terminal voltage: Vab = E + Ir
• Power supplied by charging device: P = VabI
• Power input into battery: P = E I + I2r
• Power stored in battery: E I• Power dissipated in battery: I2r
b
a
chargingdevice
ε
r
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Resistor Circuit (7)
Consider two 24V batteries with internal resistances (a) r = 4Ω, (b) r = 2Ω.
• Which setting of the switch (L/R) produces the larger power dissipation in the resistor on the side?
L R
24V
4Ω
2Ω
L R
24V
2Ω4Ω 4Ω
2Ω
(a) (b)
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Impedance Matching
A battery providing an emf E with internal resistance r is connected to an external resistor of resistance R asshown.
For what value of R does the battery deliver the maximum power to the external resistor?
• Electric current: E − Ir− IR = 0 ⇒ I =E
R + r
• Power delivered to external resistor: P = I2R =E2R
(R + r)2 =E2
rR/r
(R/r + 1)2
• Condition for maximum power: dPdR
= 0 ⇒ R = r
R
εr
I
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Resistor Circuit (5)
Consider the resistor circuit shown.
(a) Choose a current direction and use the loop rule to determine the current.(b) Name the direction of positive current (cw/ccw).(c) Find the potential difference Vab = Vb −Va.(d) Find the voltage Vcd = Vd −Vc.
12V2Ω
7Ω3Ω
4Ω 4V
c
a
b
d
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Symbols Used in Cicuit Diagrams
A
V
ε
L
C
R
C
B
E
resistor
capacitor
inductor
emf source
ammeter (connect in series)
voltmeter (connect in parallel)
diode
transistor
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Resistors Connected in Series
Find the equivalent resistance of two resistors connected in series.
• Current through resistors: I1 = I2 = I
• Voltage across resistors: V1 + V2 = V
• Equivalent resistance: R ≡ VI=
V1
I1+
V2
I2
• ⇒ R = R1 + R2
V0
V1
V2
x
V + V00V
2RR1
I I
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Resistors Connected in Parallel
Find the equivalent resistance of two resistors connected in parallel.
• Current through resistors: I1 + I2 = I
• Voltage across resistors: V1 = V2 = V
• Equivalent resistance: 1R≡ I
V=
I1
V1+
I2
V2
• ⇒ 1R
=1
R1+
1R2
V0
V = V2
x
x
V
V = VV
0
0V + V
0
1
R
R2
1I
I2
1
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Resistor Circuit (1)
Consider the two resistor circuits shown.
(a) Find the resistance R1.(b) Find the emf E1.(c) Find the resistance R2.(d) Find the emf E2.
1AR1
1Ω 2Ω
1Ω
1A
2A
6Ω
R2
ε2ε1
2A
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Resistor Circuit (2)
Consider the two resistor circuits shown.
(a) Find the resistance R1.(b) Find the current I2.(c) Find the current I3.(d) Find the resistance R4.
3A
6Ω
R1Ω1
I2
12V 12V
3Ω
3A 2Ω
I3R4
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Resistor Circuit (8)
Consider the circuit of resistors shown.
• Find the equivalent resistance Req.• Find the currents I1, . . . , I5 through each resistor and the voltages V1, . . . , V5 across each resistor.• Find the total power P dissipated in the circuit.
4Ω
6Ω
12Ω
3Ω 5Ω
12V
R = R =
R =
R =
R =1
2
3
4 5
ε =
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Kirchhoff’s Rules
Loop Rule
• When any closed-circuit loop is traversed, the algebraic sum of the changes in electric potential must bezero.
Junction Rule
• At any junction in a circuit, the sum of the incoming currents must equal the sum of the outgoing currents.
Strategy
• Use the junction rule to name all independent currents.• Use the loop rule to determine the independent currents.
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Applying the Junction Rule
In the circuit of steady currents, use the junction rule to find the unknown currents I5, . . . , I9.
Ι 8Ι
ε+
Ι
Ι Ι
−
1
2
5
3
6
79
4
I = 4A
I = 9A
I = 3A
I = 1A
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Applying Kirchhoff’s Rules
Consider the circuit shown below.
• Junction a: I1, I2 (in); I1 + I2 (out)• Junction b: I1 + I2 (in); I1, I2 (out)• Two independent currents require the use of two loops.• Loop A (ccw): 6V− (2Ω)I1 − 2V− (2Ω)I1 = 0
• Loop B (ccw): (3Ω)I2 + 1V + (2Ω)I2 − 6V = 0
• Solution: I1 = 1A, I2 = 1AI2
1
A2V 6V
2Ω 1I
2
b
aI + I
I + I
1 2
B
Ω2
2Ω I1 I2 Ω3
1V
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Resistor Circuit (11)
Consider the electric circuit shown.
• Identify all independent currents via junction rule.• Determine the independent currents via loop rule.• Find the Potential difference Vab = Vb −Va.
b
a
1Ω
4V 4V
2Ω
2V
1Ω
1Ω
1Ω
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Resistor Circuit (9)
Use Kirchhoff’s rules to find
(a) the current I,(b) the resistance R,(c) the emf E ,(d) the voltage Vab ≡ Vb −Va.
b
a
I
18V
R
1A
2Ω
ε
6A
2Ω
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Resistor Circuit (10)
Consider the electric circuit shown.
(a) Find the current through the 12V battery.(b) Find the current through the 2Ω resistor.(c) Find the total power dissipated.(d) Find the voltage Vcd ≡ Vd −Vc.(e) Find the voltage Vab ≡ Vb −Va.
a
b
3Ω
1Ωc
4V
d1Ω
2Ω12V
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Resistor Circuit (12)
Consider the electric circuit shown.
• Find the equivalent resistance Req of the circuit.• Find the total power P dissipated in the circuit.
1Ω
1Ω
2Ω
2Ω
3Ω
13V
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Resistor Circuit (3)
Consider the rsistor and capacitor circuits shown.
(a) Find the equivalent resistance Req.(b) Find the power P2, P3, P4 dissipated in each resistor.(c) Find the equivalent capacitance Ceq.(d) Find the energy U2, U3, U4 stored in each capacitor.
3Ω
2Ω 4Ω 2nF 4nF
3nF
12V12Vtsl150