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Lecture 10 – AC Circuits
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25-7 Alternating Current (Chapter 25)
Current from a battery
flows steadily in one
direction (direct current,
DC). Current from a
power plant varies
sinusoidally (alternating
current, AC).
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25-7 Alternating Current
The voltage varies sinusoidally with time:
,,
as does the current:
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25-7 Alternating Current
Multiplying the current and the voltage gives
the power:
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25-7 Alternating Current
Usually we are interested in the average power:
.
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25-7 Alternating Current
The current and voltage both have average
values of zero, so we square them, take the
average, then take the square root, yielding the
root-mean-square (rms) value:
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25-7 Alternating Current
Example 25-13: Hair dryer.
(a) Calculate the resistance and the peak current
in a 1000-W hair dryer connected to a 120-V line.
(b) What happens if it is connected to a 240-V line
in Britain?
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Chapter 30
Inductance, Electromagnetic
Oscillations, and AC Circuits
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Units of Chapter 30
•Energy Stored in a Magnetic Field
•LR Circuits
• LC Circuits with Resistance (LRC Circuits)
• AC Circuits with AC Source
•LRC Series AC Circuit
• Resonance in AC Circuits
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Energy Stored in a Magnetic Field
• Energy U stored in an inductor of
inductance L when it is carrying a current I
• U = ½ LI2
• The energy is stored in the magnetic field of
the inductor.
• Similarly, energy stored in a capacitor C when
potential difference across it is V
• U = ½ CV2
• The energy is stored in the electric field of
the capacitor.
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30-4 LR Circuits
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.
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30-4 LR Circuits
The sum of potential differences around the
loop gives
Integrating gives the current as a function
of time:
.
The time constant of an LR circuit is
.
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.
30-4 LR Circuits
If the circuit is then shorted across the
battery, the current will gradually decay away:
.
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30-4 LR Circuits
Example 30-6: An 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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30-7 AC Circuits with AC Source
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.
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30-7 AC Circuits with AC Source
The voltage across the
inductor is given by
or
.
Therefore, the current
through an inductor
lags the voltage by 90°.
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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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30-7 AC Circuits with AC Source
The voltage across the
capacitor is given by
.
Therefore, in a capacitor,
the current leads the
voltage by 90°.
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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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30-8 LRC Series AC Circuit
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.
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30-8 LRC Series AC Circuit
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.
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30-8 LRC Series AC Circuit
Some time t later,
the phasors have
rotated.
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30-8 LRC Series AC Circuit
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.
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30-8 LRC Series AC Circuit
We find from the ratio of voltage to
current that the effective resistance,
called the impedance, of the circuit
is given by
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30-8 LRC Series AC Circuit
The phase angle between the voltage and
the current is given by
or
The factor cos φ is called the
power factor of the circuit.
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Power in Inductor, Capacitor,
Resistor and AC Circuits
• Power is dissipated only by a resistance;
none is by inductance & capacitance.
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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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30-9 Resonance in AC Circuits
The rms current in an ac circuit is
Clearly, Irms depends on the frequency.
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30-9 Resonance in AC Circuits
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.
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Summary of Chapter 30
• LR circuit:
.
.
• Inductive reactance:
• Capacitive reactance:
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Summary of Chapter 30
• LRC series circuit:
.
• Resonance in LRC series circuit:
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