Ch 21 Alternating Current

Topic --- Alternating Current
21 ALTERNATING
CURRENT
21.1
Alternating
current
21.2
Root mean
square
(rms)
21.3
Resistance,
reactance
&
impedance
21.4
Power and
power
factor

(a) Define alternating current
(AC)
(b) Sketch and analyse
sinusoidal AC waveform
(c) Use sinusoidal voltage and
current equations
tII sin0 tVV sin0
21.1 ALTERNATING CURRENT

Alternating current (AC) electricity is the
type of electricity commonly used in
homes and businesses throughout the
world. While direct current (DC)
electricity flows in one direction through
a wire, AC electricity alternates its
direction in a back-and-forth motion.
The direction alternates between 50 and
60 times per second, depending on the
electrical system of the country.
AC created by an AC electric
generator, which determines
the frequency
the voltage can be readily
changed, thus making it more
suitable for long-distance
transmission than DC electricity
can employ capacitors and
inductors in electronic circuitry,
allowing for a wide range of
applications

• is defined as an electric
current which magnitude
& direction change
periodically
• Symbol:

Current
Voltage The output of an ac
generator is sinusoidal
and varies with time
locityangular veORfrequencyangular:
currentpeak:0I gepeak volta:0V
time:t
)2( f  In general,

tII o sin
tVV o sin

f
T
1


21.2 ROOT MEANC SQUARE (rms)
(a) Define root mean square
(rms) current and voltage
for AC source
(b) Use the following formula,
and
2
0
rms
I
I 
2
0
rms
V
V 

rmsI
acdc poweraveragepower 
RIRI ave
22

aveII 2

Mean or average current, Iave: the average
or mean value of current in a half-cycle
flows of current in a certain direction
  
0
2
0
ave
2II
I 
averms
II 2

The r.m.s (root mean square) current means the
square root of the average value of the current
21.2 ROOT MEAN SQUARE (rms)

• Since ac current
and
• So
and
• The root mean square (rms)
current is the effective value of
the AC
tII sin0
averms
II 2

  averms
tsinII
2
0

2
0
rms
I
I 
 
2
1
sinsin 22
 tt 

• The equation:
• Most household electricity is 240
V AC which means that Vrms is
240 V
2
0
rms
V
V 

A sinusoidal, 60.0 Hz, ac
voltage is read to be 120 V by
an ordinary voltmeter.
(a) What is the maximum value
the voltage takes on during
a cycle?
(b)b) What is the equation for
the voltage?
Solution:
(a)
(b)
2
o
rms
V
V 
)V(V rmso
2
V170
tsinVV o

tsin 120170

A voltage V= 60 sin 120πt is
applied across a 20 Ω resistor.
(a)What will an ac ammeter in
series with the resistor
read?
(b)Calculate the peak current
and mean power.
2
o
rms
V
V 
R
V
I rms
rms

)I(I rmso
2
2
0
I
Irms

 
W90(20)(2.12)2
2

 RIP rmsav
Solution:
(a)
(b)

1. An AC source
V = 500 sin t
is connected across a resistor of
250 . Calculate
(a) the rms current in the resistor,
(b) the peak current,
(c) the mean power.
2. Figure below shows a graph to
represent alternating current
passes through a resistor of 10
k. Calculate
(a) the rms current,
(b) the frequency of the AC,
(c) the mean power dissipated from
the resistor.

(a) Sketch and use phasor
diagram and sinusoidal
waveform to show the
phase relationship
between current and
voltage for a circuit
consisting of
(i) pure resistor
(ii)pure capacitor
(iii)pure inductor
(b) Use phasor diagram to
analyse voltage, current,
and impedance of series
circuit of:
(i) RL
(ii)RC
(iii)RLC
21.3 RESISTANCE, REACTANCE & IMPEDANCE

(c) Define and use:
(i) capacitive reactance,
(ii) inductive reactance,
(ii)impedance,
(iv)phase angle,
(d) Explain
graphically the
dependence of
R, XC, XL and Z
on f and relate it
to resonance
fC
XC
2
1

fLXL 2
 22
CL XXRZ 
R
XX CL 
tan

A diagram containing
phasor is called
phasor diagram
Phasor  a vector
that rotate
anticlockwise about
its axis with constant
angular velocity
Used to represent a
sinusoidally varying quantity
such as alternating current
(AC) & alternating voltage
Also being used to determine
the phase angle  the phase
difference between current
and voltage in AC circuit

Phase difference is
22

 





 tt
Note:
valuepositive
radian 
valuenegative
Leads
Lags behind
In antiphase

• The projection
of OP on the
y-axis is ON,
represents the
instantaneous
value
• Ao is the
peak value
of the
quantity
t0 TT
2
1 T2T
2
3
Ao
tAA o sin
y
ω
N
O
P
y

• It is defined by
OR
• It is a scalar quantity
and its unit is ohm ()
• In a DC circuit,
impedance likes the
resistance
rms
rms
I
V
Z 
2
0V
2
0I
0
0
I
V
Z 
V = IZ
 V = IR
 V = IX
 V = IXC
 V = IXL

Resistance, R: Opposition to current flow in purely resistive circuit
Reactance, X: Opposition to current flow resulting from
inductance or capacitance in ac circuit
Capacitive reactance, XC: Opposition of a capacitor to ac
Inductive reactance, XL: Opposition of an inductor to ac
Impedance, Z: Total opposition to ac (Resistance and reactance
combine to form impedance)

AC is defined as an electric
current which magnitude
& direction change
periodically
In general,

Root mean square current
(Irms) is defined as the
effective value of a.c. which
produces the same power as
the steady d.c. when the
current passes through the
same resistor
2
0
rms
I
I 
Root mean square voltage/
p.d (Vrms): the value of the
steady direct voltage which
when applied across a resistor,
produces the same power as
the mean (average) power
produced by the alternating
voltage across the same
resistor
2
0
rms
V
V 
A diagram containing
phasor is called
phasor diagram
Phasor  a vector
that rotate
anticlockwise about
its axis with constant
angular velocity
Impedance, Z
Resistance, R
in resistor
Reactance, X
Capacitive
reactance, XC
in capacitor, C
Inductive
reactance, XL
in inductor, L
V = IZ
 V = IR
 V = IX
 V = IXC
 V = IXL

• The circuit
• The alternating current passes through
the resistor is given by
• The alternating voltage across the
resistor VR at any instant is given by
where, V = supply voltage
tII sin0
IRVR 
00 VRI  RtI sin0 and
VtVVR  sin0
Pure Capacitor In
AC Circuit

• From figure above: I = I0 sin t
and V = V0 sin t
• Thus the phase difference is
Impedance in a pure
resistor
• From the
definition of the
impedance, hence
t0
0I
0V
0I
0V
TT
2
1 T2
T
2
3
ω
VI
0 tt 
Therefore the
current I is in
phase with the
voltage V and
constant with
time
R
I
V
I
V
Z 
0
0
rms
rms
Pure Capacitor In
AC Circuit

CIRCUIT
AC source
R
I
RV
V
PHASE DIFFERENCE
tII sin0
0
IMPEDANCE, Z
PHASOR
DIAGRAM
VI
RZ 
In pure resistor, the current I always in phase
with the voltage V and constant with time
R
I
V
I
V
Z 
0
0
rms
rms
Pure Capacitor In
AC Circuit

• The circuit
• The alternating voltage across the
capacitor VC at any instant is equal to the
supply voltage V and is given by
• The charge accumulates at the plates of
the capacitor is
AC source
CV
V
C
I
tVVVC sin0
CCVQ 
tCVQ sin0
Pure Capacitor In
AC Circuit

• The charge and current are related by
Hence the equation of AC in the capacitor is
and
OR
dt
dQ
I 
 tCV
dt
d
I sin0
 t
dt
d
CV sin0
tCV cos0 00 ICV 
tII cos0







2
sin0

tII
Pure Capacitor In
AC Circuit

• From figure above: V = V0 sin t
and I = I0 sin (t + /2)
the voltage V lags
behind the current I
by /2 radians
OR
the current I leads
the voltage V by /2
radians
t0
0I
0V
0I
0V
TT
2
1 T2
T
2
3
ω
VI
rad
2

 
22

 





 tt
Pure Capacitor In
AC Circuit

Impedance in a pure capacitor
• From the definition of the
impedance, hence
and
and
where XC:
Capacitive
(capacitative)
reactance
• Capacitive reactance is the
opposition of a capacitor to the
alternating current flows and is
defined by
• Capacitive reactance is a scalar
quantity and its unit is ohm ()
0
0
I
V
Z  00 CVI 
0
0
CV
V

CX
C
Z 

1
f 2
fC
XC
2
1

0
0
rms
rms
I
V
I
V
XC 
f0
CX
f
X C
1

Pure Capacitor In
AC Circuit

CIRCUIT
PHASE
DIFFERENCE
IMPEDANCE, Z PHASOR DIAGRAM
AC source
CV
V
C
I
tVVVC sin0







2
sin0

tII
rad
2

 
the voltage V lags behind
the current I by /2
radians.
OR
the current I leads the
voltage V by /2 radians
V
I
rad
2

 
fC
XC
2
1

f0
CX
f
X C
1

Pure Capacitor In
AC Circuit

An 8.00 μF capacitor is
connected to the terminals of
an AC generator with an rms
voltage of 150 V and a
frequency of 60.0 Hz. Find the
capacitive reactance rms
current and the peak current
in the circuit.
Solution:
fC
XC
2
1

0
0
rms
rms
I
V
I
V
XC 
Pure Capacitor In
AC Circuit

• The circuit
• The alternating current passes through the
inductor is given by
• When the AC passes through the inductor, the
back emf caused by the self induction is
produced and is given by
AC source
V
I
L
LV
tII sin0
dt
dI
LB  tI
dt
d
L sin0
tLI  cos0B 
At any instant,
the supply
voltage V
equals to the
back emf B in
the inductor but
the back emf
always oppose
the supply
voltage V
represents by
the negative
sign







2
sin0

 tLIVLB
Pure Inductor In
AC Circuit

• From figure above: I = I0 sin t
and V = V0 sin (t + /2)
In the pure inductor,
the voltage V leads
the current I by /2
radians
OR
the current I lags
behind the voltage V
by /2 radians
22



 





 tt
t0
0I
0V
0I
0V
TT
2
1 T2
T
2
3
V
I
rad
2

 
ω
Pure Inductor In
AC Circuit

Impedance in a pure inductor
• From the definition of the
impedance, hence
where XL:
inductive reactance
• Inductive reactance is the
opposition of a inductor to the
alternating current flows and is
defined by
• Inductive reactance is a scalar
quantity and its unit is ohm ()
0
0
I
V
Z  00 LIV and
0
0
I
LI 

LXLZ  
fLXL 2
f 2and
0
0
rms
rms
I
V
I
V
XL 
f0
LX
fX L
Pure Inductor In
AC Circuit

CIRCUIT
PHASE
DIFFERENCE
IMPEDANCE, Z PHASOR DIAGRAM
the voltage V leads the
current I by /2 radians.
OR
the current I lags behind
the voltage V by /2
radians
AC source
V
I
L
LV
tII sin0
tVV cos0
OR







2
sin0

tVV
rad
2

 
V
I
rad
2

 
f0
LX
fX L
fLXL 2
Pure Inductor In
AC Circuit

A coil having an inductance of 0.5 H
is connected to a 120 V, 60 Hz
power source. If the resistance of
the coil is neglected, what is the
effective current through the coil.
A 240 V supply with a frequency of
50 Hz causes a current of 3.0 A to
flow through an pure inductor.
Calculate the inductance of the
inductor.
0
0
rms
rms
I
V
I
V
XL 

Since the voltage
on a capacitor  to
the charge on it,
the current must
lead the voltage in
time & phase to
conduct charge to
the capacitor plates
and raise the
voltage
V
I
rad
2

 
When a voltage is
applied to an
inductor, it resists
the change in
current. The current
builds up more
slowly than the
voltage, lagging it in
time and phase
V
I
rad
2

 

Topic --- Alternating CurrentPure Capacitor in AC Circuit Pure Inductor in AC Circuit
V
I
rad
2

 
the voltage V leads the current I by /2
radians
OR
the current I lags behind the voltage V
by /2 radians
tII sin0







2
sin0

tVV
rad
2

  IV
tVVVC sin0







2
sin0

tII
rad
2

  IV
V
I
rad
2

 
the voltage V lags behind the current I
by /2 radians
OR
the current I leads the voltage V by
/2 radians
tII sin0
0
In pure resistor, the current I always in
phase with the voltage V and constant
with time
VI

Q1
A capacitor has a rms
current of 21 mA at a
frequency of 60 Hz when
the rms voltage across it is
14 V.
(i) What is the capacitance
of the capacitor?
(ii) If the frequency is
increased, will the current
in the capacitor increase,
decrease or stay the same?
Explain.
(iii) Calculate the rms
current in the capacitor at
a frequency of 410 Hz.
Q2
A 2 F capacitor and a
1000  resistor are
placed in series with an
alternating voltage
source of 12 V and
frequency of 50 Hz.
Calculate
(i) the current flowing,
(ii) the voltage across
the capacitor,
(iii) the phase angle of
the circuit.
Q3
A rms voltage of 12.2
V with a frequency of
1.00 kHz is applied to
a 0.290 mH inductor.
(i) What is the rms
current in the circuit?
(ii) Determine the
peak current for a
frequency of 2.50
kHz.

AC source
R
I
RV
V
CV
C
IRVR  CC IXV 
22
CR VVV 
Phasor diagram:
the current I
leads the supply
voltage V by 
radians
R
C
V
V
tan
R
XC
tan

CX
Z
R
22
CXRZ 
fC
XC
2
1

22
2 1
C
RZ


A phasor diagram in terms of
R, XC and ZI

CV
RV
V

An alternating current of
angular frequency of 1.0 x 104
rad s-1 flows through a 10 k
resistor and a 0.10 F
capacitor which are connected
in series. Calculate the rms
voltage across the capacitor if
the rms voltage across the
resistor is 20 V.
Solution:
fC
XC
2
1

0
0
rms
rms
I
V
I
V
XC 
RC Series Circuit

IRVR 
Phasor diagram:
the supply voltage V leads the
current I the by  radians
22
2 1
C
RZ


A phasor diagram in terms of
R, XL and Z
AC source
R
I
RV
V
L
LV
LL IXV 

LV
V
I RV
22
LR VVV 
R
L
V
V
tan

LX
Z
R
R
XL
tan
22
LXRZ 
fLXL 2

IRVR 
LL IXV 
CC IXV 
AC source
I
V
R
RV CV
C L
LV
I

LV
RV
V
CV
 CL VV 
 22
CL XXRIV 
R
XX CL 
tan

LX
Z
CX
 CL XX 
R
 22
CL XXRZ 
Phasor diagram:
the supply voltage V
leads the current I the
by  radians

A series RLC circuit has a resistance
of 25.0 Ω, a capacitance of 50.0 μF,
and an inductance of 0.300 H. If the
circuit is driven by a 120 V, 60 Hz
source, calculate
(a) The total impedance of the circuit
(b) The rms current in the circuit
(c) The phase angle between the
voltage and the current.
Suggested Answer:
RCL Series Circuit
• 64.9 Ω ,
1.85 A,
67.3o

• is defined as the phenomenon
that occurs when the
frequency of the applied
voltage is equal to the
frequency of the RCL series
circuit
• Since resonance in series
RLC circuit occurs at
particular frequency, so it is
used for filtering and tuning
purpose as it does not allow
unwanted oscillations that
would otherwise cause signal
distortion, noise and damage
to circuit to pass through it
• Figure below shows the variation of XC,
XL, R and Z with frequency f of the RCL
series circuit
Z
fXL 
R
f
XC
1

0 f
Z
rf
•at low
frequency,
impedance Z
is large
because 1/ωC
is large
• at high
frequency,
impedance Z
is high
because ωL is
large

• From the
graph,
Zmin at fr
• This will
happen
when
 22
CL XXRZ 
02
min  RZ
RZ min
fr : resonant frequency
R
V
Z
V
I 
min
max
• The resonant
frequency, fr of the
RCL series circuit is
given by
CL XX 
C
L


1

LC
12

 
LC
f
1
2
2
r 
LC
f
2
1
r 
• At resonance in the
RCL series circuit,
the impedance is
minimum Zmin
thus the rms current
flows in the circuit is
maximum Imax and is
given by
The resistance in
the circuit is only
came from R
CL XX 

A 200  resistor, a 0.75 H inductor and a
capacitor of capacitance C are connected in
series to an alternating source 250 V, fr = 600
Hz. Calculate
(a) the inductive reactance and capacitive
reactance when resonance is occurred
(b) the capacitance C
(c) the impedance of the circuit at resonance
(d) the current flows through the circuit at
resonance
(e) Sketch the phasor diagram.
Suggested Answer:
(a)
(b)
(c)
(d)
(e)
RCL Series Circuit


k2.83
k2.83
C
L
X
LX 
nF93.9,
2
1
x102.83 3
 C
fC
A1.25
R
V
Z
V
I rmsrms
rms
VL
VC
VR
I
 200RZ

1. Based on the RCL series
circuit in Figure above, the
rms voltages across R, L and
C are shown.
(a)With the aid of the phasor
diagram, determine the
applied voltage and the
phase angle of the circuit.
(b) Calculate
(i) the current flows in the circuit
if the resistance of the
resistor R is 26 ,
(ii) the inductance and
capacitance if the frequency
of the AC source is 50 Hz,
(iii)the resonant frequency.

Q2
• A 2 F capacitor and a 1000  resistor are placed in series with an
alternating voltage source of 12 V and frequency of 50 Hz. Calculate
• (a) the current flowing,
• (b) the voltage across the capacitor,
• (c) the phase angle of the circuit.
Q3
• An AC current of angular frequency of 1.0  104 rad s1 flows
through a 10 k resistor and a 0.10 F capacitor which are
connected in series. Calculate the rms voltage across the capacitor if
the rms voltage across the resistor is 20 V.
ANS: 2.0 V

Q4A 200  resistor, a 0.75 H
inductor and a capacitor of
capacitance C are connected in
series to an alternating source
250 V, 600 Hz. Calculate
(a) the inductive reactance and
capacitive reactance when
resonance is occurred.
(b) the capacitance C.
(c) the impedance of the circuit at
resonance.
(d) the current flows through the
circuit at resonance. Sketch the
phasor diagram of the circuit.
ANS: 2.83 k, 2.83 k; 93.8 nF;
200 ; 1.25 A
Q5
A capacitor of capacitance C, a coil
of inductance L, a resistor of
resistance R and a lamp of
negligible resistance are placed in
series with alternating voltage V. Its
frequency f is varied from a low to a
high value while the magnitude of V
is kept constant.
(a) Describe and explain how the
brightness of the lamp varies.
(b) If V=0.01 V, C =0.4 F, L =0.4 H,
R = 10  and the circuit at
resonance, calculate
(i) the resonant frequency,
(ii) the maximum rms current,
(iii) the voltage across the
capacitor.
• (Advanced Level Physics,7th edition, Nelkon
& Parker, Q2, p.423)
• ANS: 400 Hz; 0.001 A; 1 V

V
V V
22
LR VVV 
22
LXRIV 
22
LXRZ 
IZV 

A circuit is made up of a 3200
pF capacitor connected in
series to a 30 H coil of
resistance 4 . Calculate
(i) impedance at frequency
30 kHz.
(ii)resonant frequency.
Solution:
C = 3200  10-12 F; L = 3010-6
H; R = 4 
(i) Given f = 30103 Hz, The
reactance of capacitor and
inductor are
PSPM 2009/ 10: Q12(c):
fC
XC
2
1

  123
10320010302
1




 1066.1 3
CX

(i) and
Therefore the impedance is given
by
(ii)Apply:
fLXL 2
  63
103010302 
 
 22
CL XXRZ 
   232
1066.166.54 
1654Z
 66.5LX
  126
10320010302
1




Hz1014.5 5
rf
LC
fr
2
1


Apply
(i) average power,
(ii)instantaneous power,
(iii)power factor,
in AC circuit consisting of R, RC, RL and RCL in
series.
21.4 POWER & POWER FACTOR
• Power factor is
a way of
measuring how
efficiently
electrical
power is being
used within a
facility's
electrical
system
cosVIP rmsrms
av
IVP 
IV
P
P
P av
a
r
cos 

• In an ac circuit , the power
is only dissipated by a
resistance, none is
dissipated by inductance or
capacitance
• From the phasor diagram
of the RCL series circuit
RIIVP R
2
av 

ω
LV
I RV
V
CV
 CL VV 
• We get
• Then
and
cosVVR 
V
VR
cos
cosav IVP  IZV 
r
2
av cos PZIP  
where cos  is
called the
power factor of
the AC circuit,
Pr is the
average real
power and I2Z
is called the
apparent
power

• Power factor is defined as
• From
• the power factor also can be
calculated by using the equation
below
• When  = 0o (cos  =+1) ,the circuit
is completely resistive or when the
circuit is in resonance (RCL)
• When  = +90o (cos  = 0), the
circuit is completely inductive
• When  = -90o (cos  =0), the circuit
is completely capacitive
a
r
2
r
cos
P
P
ZI
P

IZ
IR
V
VR
cos
Z
R
cos
 22
CL XXRIV 

An oscillator set for 500 Hz puts out a sinusoidal voltage of 100 V
effective. A 24.0 Ω resistor, a 10.0μF capacitor, and a 50.0 mH
inductor in series are wired across the terminals of the oscillator.
(a)What will an ammeter in the circuit read?
(b)What will a voltmeter read across each element?
(c)What is the real power dissipated in the circuit?
(d)Calculate the power supply
(e)Find the power factor
(f) What is the phase angle?
 R1
Z
V
I rms
rms

CC
LL
R
IXV
IXV
IRV



cosVIP rmsave rmsrmsplysup
VIP 
Z
R
cos  





 
Z
R
cos 1


A 100 F capacitor, a 4.0 H inductor and a 35  resistor are
connected in series with an alternating source given by the
equation. V = 520 sin 100t. Calculate:
(a)the frequency of the source,
(b)the capacitive reactance and inductive reactance,
(c)the impedance of the circuit,
(d)the peak current in the circuit,
(e)the phase angle,
(f) the power factor of the circuit.

Suggested Answer:
By comparing
V = 520 sin 100t
to the, V = V0 sin t
Thus
V0 = 520V,  = 100 rad s-1
(a) The frequency of AC source is
given by
(b) The capacitive reactance is
and the inductive reactance is
f 2
Hz9.15f
f2100 
fC
XC
2
1

100CX
  6
101009.152
1




CX
fLXL 2
 400LX
  0.49.152

(c) The impedance of the circuit is
(d) The peak current in the circuit is
(e) The phase angle between the
current and the supply voltage is
OR
f. The power factor of the circuit is
given by
 22
CL XXRZ 
   22
10040035 
 302Z
ZIV 00 
 302520 0I
A72.10 I
R
XX CL 
tan





 
 
35
100400
tan 1
rad45.1





 
 
R
XX CL1
tan

3.83
cosfactorpower 

383cos .
117.0factorpower 

Q1
A 22.5 mH inductor, a
105  resistor and a
32.3 F capacitor are
connected in series to
the alternating source
240 V, 50 Hz.
(a) Sketch the phasor
diagram for the circuit
(b) Calculate the power
factor of the circuit
(c) Determine the
average power
consumed by the
circuit.
ANS: 0.755, 313 W
Q2
A coil having inductance 0.14
H and resistance of 12  is
connected to an alternating
source 110 V, 25 Hz.
Calculate
(a) the rms current flows in
the coil
(b) the phase angle between
the current and supply
voltage
(c) the power factor of the
circuit
(d) the average power loss in
the coil.
ANS: 4.4 A, 61.3o , 0.48, 0.23
kW
Q3
A series RCL circuit
contains a 5.10 μF
capacitor and a generator
whose voltage is 11.0 V.
At a resonant frequency
of 1.30 kHz the power
dissipated in the circuit is
25.0 W. Calculate
(a) the inductance
(b) the resistance
(c) the power factor when
the generator frequency
is 2.31 kHz.
ANS: 2.94 x 10-3 H , 4.84
Ω , 0.163

Q4
• An RLC circuit has a resistance of 105 , an inductance of 85.0 mH and a
capacitance of 13.2 F.
• What is the power factor of the circuit if it is connected to a 125 Hz AC
generator?
• Will the power factor increase, decrease or stay the same if the resistance is
increased? Explain.
• (Physics, 3rd edition, James S. Walker, Q47, p.834)
• ANS: 0.962; U think
Q5
• A 1.15 k resistor and a 505 mH inductor are connected in series to a 14.2 V,1250 Hz AC
generator.
• What is the rms current in the circuit?
• What is the capacitance’s value must be inserted in series with the resistor and inductor to
reduce the rms current to half of the value in part (a)?
• (Physics, 3rd edition, James S. Walker, Q69, p.835)
• ANS: 3.44 mA, 10.5 nF

G E O M E T R I C A L
O P T I C S

Ch 21 Alternating Current

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