inductance and ac circuits
DESCRIPTION
Inductance and AC Circuits. Mutual Inductance Self-Inductance Energy Stored in a Magnetic Field LR Circuits LC Circuits and Electromagnetic Oscillations LC Circuits with Resistance ( LRC Circuits) AC Circuits with AC Source. LRC Series AC Circuit Resonance in AC Circuits - PowerPoint PPT PresentationTRANSCRIPT
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Inductance and AC Circuits
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• Mutual Inductance
• Self-Inductance
• Energy Stored in a Magnetic Field
• LR Circuits
• LC Circuits and Electromagnetic Oscillations
• LC Circuits with Resistance (LRC Circuits)
• AC Circuits with AC Source
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• LRC Series AC Circuit
• Resonance in AC Circuits
• Impedance Matching
• Three-Phase AC
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Homework
• Ch. 32– Exer. 3, 4, 5, 6, 9, 10, 11, 12,
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Inductance
• Induced emf in one circuit due to changes in the magnetic field produced by the second circuit is called mutual induction.
• Induced emf in one circuit associated with changes in its own magnetic field is called self-induction.
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Inductance
2221
222122
2221222
dt
dN
NNdt
Nd
Unit of inductance: the henry, H:
1 H = 1 V·s/A = 1 Ω·s.
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Mutual inductance: magnetic flux through coil2 due to current in coil 1
Induced emf due to mutual induction:
Mutual Inductance
121212 IMN
dt
dIM
dt
dN 1
2121
221
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Mutual InductanceSolenoid and coil.
A long thin solenoid of length l and cross-sectional area A contains N1 closely packed turns of wire. Wrapped around it is an insulated coil of N2 turns. Assume all the flux from coil 1 (the solenoid) passes through coil 2, and calculate the mutual inductance.
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Mutual Inductance
Reversing the coils.
How would the previous example change if the coil with turns was inside the solenoid rather than outside the solenoid?
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A changing current in a coil will also induce an emf in itself:
Self-inductance: magnetic flux through the coil due to the current in the coil itself:
Self-Inductance
2222 LIN
dt
dIL
dt
dN 222
222
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Self-Inductance
Solenoid inductance.
(a) Determine a formula for the self-inductance L of a tightly wrapped and long solenoid containing N turns of wire in its length l and whose cross-sectional area is A.
(b) Calculate the value of L if N = 100, l = 5.0 cm, A = 0.30 cm2, and the solenoid is air filled.
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Self-Inductance
Direction of emf in inductor.
Current passes through a coil from left to right as shown. (a) If the current is increasing with time, in which direction is the induced emf? (b) If the current is decreasing in time, what then is the direction of the induced emf?
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Self-Inductance
Coaxial cable inductance.
Determine the inductance per unit length of a coaxial cable whose inner conductor has a radius r1 and the outer conductor has a radius r2. Assume the conductors are thin hollow tubes so there is no magnetic field within the inner conductor, and the magnetic field inside both thin conductors can be ignored. The conductors carry equal currents I in opposite directions.
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Homework
• Ch. 32– Exer. 17, 18, 19, 23, 24, 25,– Exer. 28, 29, 30, 34, 35, 36,– Exer. 37, 38, 39, 40,– Prob. 5, 10,
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A circuit consisting of an inductor and a resistor will begin with most of the voltage drop across the inductor, as the current is changing rapidly. With time, the current will increase less and less, until all the voltage is across the resistor.
LR Circuits
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LR Circuits
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flow tostartscurrent the
and connected isbattery the,0At
0
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dt
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VVdt
dIL
dtdI
t
L
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LR Circuits
RVyyIt
tL
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L
R
y
y
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R
y
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Ly
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LR Circuits
R
Lτ
tR
V
tII
constant with time
exp1
exp1
Therefore,
0
0
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If the circuit is then shorted across the battery, the current will gradually decay away:
LR Circuits
.
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LR Circuits
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decrease tostartscurrent theand
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VVdt
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dtdI
t
L
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R
L
tIItL
R
I
I
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IIt
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I
dIdt
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constant timeagain with
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00
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LR CircuitsAn LR circuit.
At t = 0, a 12.0-V battery is connected in series with a 220-mH inductor and a total of 30-Ω resistance, as shown. (a) What is the current at t = 0? (b) What is the time constant? (c) What is the maximum current? (d) How long will it take the current to reach half its maximum possible value? (e) At this instant, at what rate is energy being delivered by the battery, and (f) at what rate is energy being stored in the inductor’s magnetic field?
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Just as we saw that energy can be stored in an electric field, energy can be stored in a magnetic field as well, in an inductor, for example.
Analysis shows that the energy density of the field is given by
Energy Density of a Magnetic Field
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Energy Stored in an Inductor
dt
diLiR
The equation governs the LR circuit is
Multiplying each term by the current i leads to
dt
diLiRii 2
inductor. thefrom dissipatedpower theis
battery. by the deliveredpower total theis 2Ri
i
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Energy Stored in an Inductor
dt
diLi
dt
dUL
Therefore, the third term represents the rate at which the energy is stored in the inductor
The total energy stored from i=0 to i=I is
2
0L 2
1LI
dt
diLiUI
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Energy Density of a Magnetic Field
AB
LIU0
22
L 22
1
The self-inductance of a solenoid is L=μ0nA2l. The magnetic field inside it is B=μ0nI. The energy stored thus is
Since Al is the volume of the solenoid, the energy per volume is
0
2
B 2B
u
This is the energy density of a magnetic field in free space.
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LC Circuits and Electromagnetic Oscillations
An LC circuit is a charged capacitor shorted through an inductor.
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Electromagnetic Oscillations
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The current causes the charge in the capacitor to decreases so I=-dQ/dt. Thus the differential equation becomes
LC Circuits
0dt
dIL
C
Q
Across the capacitor, the voltage is raised by Q/C. As the current passes through the inductor, the induced emf is –L(dI/dt). The Kirchhof’s loop rule gives
02
2
LC
Q
dt
Qd
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The charge therefore oscillates with a natural angular frequency
LC Circuits and Electromagnetic Oscillations
022
2
xdt
xd
The equation describing LC circuits has the same form as the SHO equation:
.LC
1
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Electromagnetic Oscillations
tQQ cos0
The charge varies as
The current is sinusoidal as well:
2cossin
sin
00
0
tItI
tQdt
dQI
Remark: When Q=Q0 at t=t0, we have φ=0.
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LC Circuits and Electromagnetic Oscillations
The charge and current are both sinusoidal, but with different phases.
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LC Circuits and Electromagnetic Oscillations
The total energy in the circuit is constant; it oscillates between the capacitor and the inductor:
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LC Circuits and Electromagnetic Oscillations
LC circuit.
A 1200-pF capacitor is fully charged by a 500-V dc power supply. It is disconnected from the power supply and is connected, at t = 0, to a 75-mH inductor. Determine: (a) the initial charge on the capacitor; (b) the maximum current; (c) the frequency f and period T of oscillation; and (d) the total energy oscillating in the system.
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LRC Circuits
Any real (nonsuperconducting) circuit will have resistance.
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LRC Circuits
Adding a resistor in an LC circuit is equivalent to adding –IR in the equation of LC oscillation
0dt
dILIR
C
Q
Initially Q=Q0, and the switch is closed at t=0, the current is I=-dQ/dt. The differential equation becomes
02
2
LC
Q
dt
dQ
L
R
dt
Qd
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LRC Circuits
The equation describing LRC circuits now has the same form as the equation for the damped oscillation:
022
2
xdt
dx
dt
xd
The solution to LRC circuits therefore is
t
L
RtQQ cos
2exp0
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LRC Circuits
where ω02=1/LC.
The system will be underdamped for R2 < 4L/C, and overdamped for R2 > 4L/C. Critical damping will occur when R2 = 4L/C.
The damped angular frequency is
220 2
L
R
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LRC Circuits
This figure shows the three cases of underdamping, overdamping, and critical damping.
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LRC Circuits
Damped oscillations.
At t = 0, a 40-mH inductor is placed in series with a resistance R = 3.0 Ω and a charged capacitor C = 4.8 μF. (a) Show that this circuit will oscillate. (b) Determine the frequency. (c) What is the time required for the charge amplitude to drop to half its starting value? (d) What value of R will make the circuit nonoscillating?
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Resistors, capacitors, and inductors have different phase relationships between current and voltage when placed in an ac circuit.
The current through a resistor is in phase with the voltage.
30-7 AC Circuits with AC Source
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Therefore, the current through an inductor lags the voltage by 90°.
30-7 AC Circuits with AC Source
The voltage across the inductor is given by
or
.
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30-7 AC Circuits with AC Source
The voltage across the inductor is related to the current through it:
The quantity XL is called the inductive reactance, and has units of ohms:
.
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30-7 AC Circuits with AC Source
Example 30-9: Reactance of a coil.
A coil has a resistance R = 1.00 Ω and an inductance of 0.300 H. Determine the current in the coil if (a) 120-V dc is applied to it, and (b) 120-V ac (rms) at 60.0 Hz is applied.
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Therefore, in a capacitor, the current leads the voltage by 90°.
30-7 AC Circuits with AC Source
The voltage across the capacitor is given by
.
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30-7 AC Circuits with AC Source
The voltage across the capacitor is related to the current through it:
The quantity XC is called the capacitive reactance, and (just like the inductive reactance) has units of ohms:
.
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30-7 AC Circuits with AC Source
Example 30-10: Capacitor reactance.
What is the rms current in the circuit shown if C = 1.0 μF and Vrms = 120 V? Calculate (a) for f = 60 Hz and then (b) for f = 6.0 x 105 Hz.
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30-7 AC Circuits with AC Source
This figure shows a high-pass filter (allows an ac signal to pass but blocks a dc voltage) and a low-pass filter (allows a dc voltage to be maintained but blocks higher-frequency fluctuations).
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Analyzing the LRC series AC circuit is complicated, as the voltages are not in phase – this means we cannot simply add them. Furthermore, the reactances depend on the frequency.
30-8 LRC Series AC Circuit
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We calculate the voltage (and current) using what are called phasors – these are vectors representing the individual voltages.
Here, at t = 0, the current and voltage are both at a maximum. As time goes on, the phasors will rotate counterclockwise.
30-8 LRC Series AC Circuit
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Some time t later, the phasors have rotated.
30-8 LRC Series AC Circuit
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The voltages across each device are given by the x-component of each, and the current by its x-component. The current is the same throughout the circuit.
30-8 LRC Series AC Circuit
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We find from the ratio of voltage to current that the effective resistance, called the impedance, of the circuit is given by
30-8 LRC Series AC Circuit
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30-8 LRC Series AC Circuit
The phase angle between the voltage and the current is given by
The factor cos φ is called the power factor of the circuit.
or
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30-8 LRC Series AC Circuit
Example 30-11: LRC circuit.
Suppose R = 25.0 Ω, L = 30.0 mH, and C = 12.0 μF, and they are connected in series to a 90.0-V ac (rms) 500-Hz source. Calculate (a) the current in the circuit, (b) the voltmeter readings (rms) across each element, (c) the phase angle , and (d) the power dissipated in the circuit.
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The rms current in an ac circuit is
Clearly, Irms depends on the frequency.
30-9 Resonance in AC Circuits
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We see that Irms will be a maximum when XC = XL; the frequency at which this occurs is
f0 = ω0/2π is called the resonant frequency.
30-9 Resonance in AC Circuits
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30-10 Impedance MatchingWhen one electrical circuit is connected to another, maximum power is transmitted when the output impedance of the first equals the input impedance of the second.
The power delivered to the circuit will be a minimum when dP/dt = 0; this occurs when R1 = R2.
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30-11 Three-Phase AC
Transmission lines usually transmit three-phase ac power, with the phases being separated by 120°. This makes the power flow much smoother than if a single phase were used.
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30-11 Three-Phase AC
Example 30-12: Three-phase circuit.
In a three-phase circuit, 266 V rms exists between line 1 and ground. What is the rms voltage between lines 2 and 3?
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• Mutual inductance:
• Energy density stored in magnetic field:
Summary of Chapter 30
• Self-inductance:
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Summary of Chapter 30• LR circuit:
• LC circuit:
.
.
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Summary of Chapter 30
• LRC series circuit:
• Resonance in LRC series circuit:
.