Download Chapter 21: Alternating Currents

Chapter 21: Alternating Currents
•Sinusoidal Voltages and Currents
•Capacitors, Resistors, and Inductors in AC Circuits
•Series RLC Circuits
•Resonance
•AC to DC Conversion
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Sinusoidal Currents and Voltage
A power supply can be set to give an EMF of form:
ε (t ) = ε 0 sin ωt
This EMF is time dependent, has an amplitude ε0, and
varies with angular frequency ω.
2
ω = 2πf
angular
frequency
in rads/sec
frequency in
cycles/sec or Hz
The current in a resistor is still given by Ohm’s Law:
I (t ) =
ε (t )
R
=
ε0
R
sin ωt = I 0 sin ωt
The current has an amplitude of I0=ε0/R.
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The instantaneous power dissipated in a resistor will be:
P = I (t )VR (t )
= (I 0 sin ωt )(ε 0 sin ωt ) = I 0ε 0 sin 2 ωt
The power dissipated depends on t (where in the cycle the
current/voltage are).
4
What is the average power dissipated by a resistor in one
cycle?
The average value sin2ωt over one cycle is 1/2.
1
The average power is Pav = I 0ε 0 .
2
5
What are the averages of V(t) and I(t) over one cycle?
The “problem” here is that the average value of sin ωt over
one complete cycle is zero! This is not a useful way to
characterize the quantities V(t) and I(t).
To fix this problem we use the root mean square (rms) as
the characteristic value over one cycle.
I rms
I0
=
2
and ε rms =
ε0
2
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In terms of rms quantities, the power dissipated by a resistor
can be written as:
I0 ε 0
1
Pav = I 0ε 0 =
2
2 2
2
= I rmsε rms = I rms
R=
2
ε rms
R
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Example (text problem 21.4): A circuit breaker trips when the
rms current exceeds 20.0 A. How many 100.0 W light bulbs
can run on this circuit without tripping the breaker? (The
voltage is 120 V rms.)
Each light bulb draws a current given by:
Pav = I rmsε rms
100 Watts = I rms (120 V )
I rms = 0.83 Amps
If 20 amps is the maximum current, and 0.83 amps is the
current drawn per light bulb, then you can run 24 light bulbs
without tripping the breaker.
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Example (text problem 21.10): A hair dryer has a power
rating of 1200 W at 120 V rms. Assume the hair dryer is the
only resistance in the circuit.
(a) What is the resistance of the heating element?
Pav =
ε 2 rms
R
2
(
120 V )
1200 Watts =
R
R = 12 Ω
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Example continued:
(b) What is the rms current drawn by the hair dryer?
Pav = I rmsε rms
1200 Watts = I rms (120 V )
I rms = 10 Amps
(c) What is the maximum instantaneous power that the
resistance must withstand?
P = I 0ε 0 sin ωt ⇒ Pmax = I 0ε 0
2
1
Pav = I 0ε 0
2
Pmax = 2Pav = 2400 Watts
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Capacitors, Resistors and
Inductors in AC circuits
For a capacitor:
Q(t ) = CVC (t )
ΔQ(t )
⎛ ΔVC (t ) ⎞
= C⎜
In the circuit: I (t ) =
⎟
Δt
⎝ Δt ⎠
Slope of the
plot V(t) vs. t
11
12
The current in the circuit and the voltage drop across the
capacitor are 1/4 cycle out of phase. Here the current leads
the voltage by 1/4 cycle.
Here it is true that VC∝I. The equality is Vc = IXC where XC
is called capacitive reactance. (Think Ohm’s Law!)
1
XC =
ωC
Reactance has
units of ohms.
13
For a resistor in an AC
circuit,
V (t ) = I (t ) R.
The voltage and current will be in phase with each other.
14
For an inductor in an AC circuit:
⎛ ΔI (t ) ⎞
VL = L ⎜
⎟
⎝ Δt ⎠
Slope of an
I(t) vs. t plot
Also, VL = IXL where the inductive reactance is:
X L = ωL
15
The current in the circuit and the voltage drop across the
inductor are 1/4 cycle out of phase. Here the current lags
the voltage by 1/4 cycle.
16
Plot of I(t), V(t), and P(t) for a capacitor.
The average power over one cycle is zero.
An ideal capacitor dissipates no energy.
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A similar result is found for inductors; no energy is dissipated
by an ideal inductor.
18
Series RLC Circuits
19
Applying Kirchhoff’s loop rule:
ε (t ) − VL (t ) − VR (t ) − VC (t ) = 0
ε (t ) = ε 0 sin (ωt + φ )
π⎞
π⎞
⎛
⎛
= VL sin ⎜ ωt + ⎟ + VR sin (ωt ) + VC sin ⎜ ωt − ⎟
2⎠
2⎠
⎝
⎝
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To find the amplitude (ε0) and phase (φ) of the total voltage
we add VL, VR, and VC together by using phasors.
y
ε 0 = V + (VL − VC )
2
2
R
=
VL
ε0
(IR ) + (IX L − IX C )
2
2
= I R + (X L − X C )
2
VR
X
2
= IZ
VC
Z is called impedance.
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y
The phase angle between the
current in the circuit and the input
voltage is:
VL
ε0
VL − VC X L − X C
=
tan φ =
VR
R
φ
VR
VC
X
VR
R
cos φ =
=
ε0 Z
φ>0 when XL> XC and the voltage leads the current
(shown above).
φ<0 when XL< XC and the voltage lags the current.
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Example (text problem 21.79): In an RLC circuit these three
elements are connected in series: a resistor of 20.0 Ω, a 35.0
mH inductor, and a 50.0 μF capacitor. The AC source has an
rms voltage of 100.0 V and an angular frequency of 1.0×103
rad/sec. Find…
(a) The reactances of the capacitor and the inductor.
X L = ωL = 35.0 Ω
1
XC =
= 20.0 Ω
ωC
(b) The impedance.
Z = R + ( X L − X C ) = 25.0 Ω
2
2
23
Example continued:
(c) The rms current:
ε rms = I rms Z
ε rms 100.0 V
I rms =
Z
=
25.0 Ω
= 4.00 Amps
(d) The current amplitude:
I rms
I0
=
2
I 0 = 2 I rms = 5.66 Amps
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Example continued:
(e) The phase angle:
X L − X C 35Ω − 20Ω
=
= 0.75
tan φ =
R
20Ω
φ = tan −1 (0.75) = 0.644 rads
(Or 37°)
(f) The rms voltages across each circuit element:
Vrms, R = I rms R = 80.0 V
Vrms, L = I rms X L = 140 V
Vrms,C = I rms X C = 80.0 V
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Example continued:
(g) Does the current lead or lag the voltage?
Since XL>XC, φ is a positive angle. The voltage leads
the current.
(h) Draw a phasor diagram.
y
VL
εrms
φ
VC
VR
X
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The power dissipated by a resistor is:
Pav = I rmsε rms,R = I rmsε rms cos φ
where cosφ is called the power factor (compare to slide 7;
Why is there a difference?).
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Resonance in RLC Circuits
A plot of I vs.
ω for a series
RLC circuit
has a peak at
ω = ω0.
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The peak occurs at the resonant frequency for the circuit.
I=
ε
Z
=
ε
R2 + (X L − X C )
2
The current will be a maximum when Z is a minimum. This
occurs when XL = XC (or when Z=R).
X L = XC
ω0 L =
1
ω0C
ω0 =
1
LC
This is the resonance
frequency for the circuit.
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At resonance:
X L − XC
tan φ =
=0
R
R
cos φ = = 1
R
The phase angle is 0; the voltage and the current are in
phase. The current in the circuit is a maximum as is the
power dissipated by the resistor.
30
Converting AC to DC; Filters
A diode is a circuit element that allows current to pass
through in one direction, but not the other.
31
The plot shows the voltage drop across the
resistor. During half a cycle, it is zero.
Putting a capacitor in the circuit “smoothes” out VR,
producing a nearly constant voltage drop (a DC voltage).
32
A capacitor may be used as a filter.
Low-pass filter. When
XC << R (ω is large) the
output voltage will be
small compared to the
input voltage.
When XC >> R (ω is small), the output voltage will be
comparable to the input voltage.
This circuit will allow low frequency signals to pass through
while filtering out high frequency signals.
33
A high-pass filter. This will allow high frequency signals
to pass through while filtering out low frequency signals.
34
Summary
•Difference Between Instantaneous, Average, and rms
Values
• Power Dissipation by R, L, and C
•Reactance for R, L, and C
•Impedance and Phase Angle
•Resonance in an RLC Circuit
•Diodes
•High- and Low-Pass Filters
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