AC Source and RLC Circuits
X L = 2π fL
Inductive reactance
X C = 1/ 2π fC
Capacitive reactance
Z = R + ( X L − XC )
2
I max =
ε max
Z
X − XC
tan φ = L
R
2
Total impedance
Maximum current
Phase angle
PHY2054: Chapter 21
1
Pictorial Understanding of Reactance
Z = R2 + ( X L − X C )
2
X L − XC
tan φ =
R
R
cos φ =
Z
PHY2054: Chapter 21
2
Summary of Circuit Elements, Impedance, Phase Angles
Z = R2 + ( X L − X C )
2
PHY2054: Chapter 21
X L − XC
tan φ =
R
3
Quiz
ÎThree
identical EMF sources are hooked to a single circuit
element, a resistor, a capacitor, or an inductor. The
current amplitude is then measured as a function of
frequency. Which one of the following curves corresponds
to an inductive circuit?
‹ (1)
a
‹ (2) b
‹ (3) c
‹ (4) Can’t tell without more info
a
Imax
b
c
X L = 2π fL
I max = ε max / X L
f
For inductor, higher frequency gives higher
reactance, therefore lower current
PHY2054: Chapter 21
4
RLC Example 1
ÎBelow
are shown the driving emf and current vs time of
an RLC circuit. We can conclude the following
‹ Current
“leads” the driving emf (φ<0)
‹ Circuit is capacitive (XC > XL)
I
ε
t
PHY2054: Chapter 21
5
RLC Example 2
ÎR
= 200Ω, C = 15μF, L = 230mH, εmax = 36v, f = 60 Hz
‹ Resonant
frequency f 0
= 1/ 2π
( 0.230 ) (15 ×10−6 ) = 85.6 Hz
‹
X L = 2π × 60 × 0.23 = 86.7Ω
‹
X C = 1/ 2π × 60 × 15 × 10−6 = 177Ω
‹
Z = 200 + ( 86.7 − 177 ) = 219Ω
‹
I max = ε max / Z = 36 / 219 = 0.164 A
‹
(
)
2
2
−1 ⎛ 86.7 − 177 ⎞
φ = tan ⎜
⎝
XC > XL
Capacitive circuit
200
⎟ = −24.3°
⎠
PHY2054: Chapter 21
Current leads emf
(as expected)
6
Imax vs Frequency and Resonance
ÎCircuit
parameters: C = 2.5μF, L = 4mH, εmax = 10v
‹ f0
= 1 / 2π(LC)1/2 = 1590 Hz
‹ Plot Imax vs f
I max = 10 / R + ( 2π fL − 1/ 2π fC )
2
2
R = 5Ω
R = 10Ω
Imax
R = 20Ω
Resonance
f = f0
f / f0
PHY2054: Chapter 21
7
Power in AC Circuits
2
P
=
I
ÎRecall power formula ave
rms R
ÎRewrite
using I rms =
ε rms
Z
ε rms
Z
I rms R = ε rms I rms cos φ
R
cos φ =
Z
Pave = ε rms I rms cos φ
Îcosφ
Pave =
I rms = I max / 2
is the “power factor”
maximize power delivered to circuit ⇒ make φ close to zero
‹ Max power delivered to load happens at resonance
‹ E.g., too much inductive reactance (XL) can be cancelled by
increasing XC (e.g., circuits with large motors)
‹ To
PHY2054: Chapter 21
8
Power Example 1
ÎR
= 200Ω, XC = 150Ω, XL = 80Ω, εrms = 120v, f = 60 Hz
‹
Z = 200 + ( 80 − 150 ) = 211.9Ω
‹
I rms = ε rms / Z = 120 / 211.9 = 0.566 A
‹
2
2
−1 ⎛ 80 − 150 ⎞
φ = tan ⎜
⎟ = −19.3°
⎠
Current leads emf
Capacitive circuit
‹
⎝ 200
cos φ = 0.944
‹
Pave = ε rms I rms cos φ = 120 × 0.566 × 0.944 = 64.1W
‹
Same
2
Pave = I rms
R = 0.5662 × 200 = 64.1W
PHY2054: Chapter 21
9
Power Example 1 (cont)
ÎR
= 200Ω, XC = 150Ω, XL = 80Ω, εrms = 120v, f = 60 Hz
ÎHow
much capacitance must be added to maximize the
power in the circuit (and thus bring it into resonance)?
‹ Want
XC = XL to minimize Z, so must decrease XC
‹
X C = 150Ω = 1/ 2π fC
C = 17.7μF
‹
X C new = X L = 80Ω
Cnew = 33.2μF
‹
So we must add 15.5μF capacitance to maximize power
PHY2054: Chapter 21
10
Power vs Frequency and Resonance
ÎCircuit
parameters: C = 2.5μF, L = 4mH, εmax = 10v
‹ f0
= 1 / 2π(LC)1/2 = 1590 Hz
‹ Plot Pave vs f for different R values
R = 2Ω
R = 5Ω
Pave
R = 10Ω
Resonance
R = 20Ω
f = f0
f / f0
PHY2054: Chapter 21
11
Quiz
ÎA
generator produces current at a frequency of 60 Hz with
peak voltage and current amplitudes of 100V and 10A,
respectively. What is the average power produced if they
are in phase?
‹ (1)
‹ (2)
‹ (3)
‹ (4)
‹ (5)
1000 W
707 W
1414 W
500 W
250 W
Pave = 12 ε peak I peak = ε rms I rms
PHY2054: Chapter 21
12
Quiz
ÎThe
figure shows the current and emf of a series RLC
circuit. To increase the rate at which power is delivered to
the resistive load, which option should be taken?
‹ (1)
Increase R
‹ (2) Decrease L
‹ (3) Increase L
‹ (4) Increase C
X L − XC
tan φ =
R
Current lags applied emf (φ > 0), thus circuit is inductive. Either
(1) Reduce XL by decreasing L or
(2) Cancel XL by increasing XC (decrease C).
PHY2054: Chapter 21
13
Example: LR Circuit
frequency EMF source with εm=6V connected to a
resistor and inductor. R=80Ω and L=40mH.
ÎVariable
‹ At
what frequency f does VR = VL?
X L = 2π fL = R ⇒ f = 318Hz
‹ At
that frequency, what is phase angle φ?
tan φ = X L / R = 1 ⇒ φ = 45°
‹ What
is the current amplitude?
I max = ε max / 802 + 802 = 6 /113 = 0.053A
‹ What
is the rms current?
I rms = I max / 2 = 0.037 A
PHY2054: Chapter 21
14
Transformers
ÎPurpose:
change alternating (AC) voltage to a bigger (or
smaller) value
Input AC voltage
in the “primary”
turns produces a flux
Vp = N p
Changing flux in
“secondary” turns
induces an emf
ΔΦ B
Vs = N s
Δt
ΔΦ B
Δt
Ns
Vs = V p
Np
PHY2054: Chapter 21
15
Transformers
ÎNothing
comes for free, however!
‹ Increase
in voltage comes at the cost of current.
‹ Output power cannot exceed input power!
‹ power in = power out
‹ (Losses usually account for 10-20%)
i pVp = isVs
is V p N p
=
=
i p Vs N s
PHY2054: Chapter 21
16
Transformers: Sample Problem
ÎA
transformer has 330 primary turns and 1240 secondary
turns. The input voltage is 120 V and the output current
is 15.0 A. What is the output voltage and input current?
Ns
⎛ 1240 ⎞
Vs = V p
= 120 ⎜
⎟ = 451V
Np
⎝ 330 ⎠
i pV p = isVs
“Step-up”
transformer
Vs
⎛ 451 ⎞
i p = is
= 15 ⎜
⎟ = 56.4 A
Vp
⎝ 120 ⎠
PHY2054: Chapter 21
17
Transformers
¾ This is how first experiment by
Faraday was done
¾ He only got a deflection of the
galvanometer when the switch
is opened or closed
¾ Steady current does not make
induced emf.
PHY2054: Chapter 21
18
Applications
Microphone
Tape recorder
PHY2054: Chapter 21
19
ConcepTest: Power lines
ÎAt
large distances, the resistance of power lines becomes
significant. To transmit maximum power, is it better to
transmit (high V, low i) or (high i, low V)?
‹ (1)
high V, low i
‹ (2) low V, high i
‹ (3) makes no difference
Power loss is i2R
PHY2054: Chapter 21
20
Electric Power Transmission
i2R: 20x smaller current ⇒ 400x smaller power loss
PHY2054: Chapter 21
21