Download Chapter 17 Alternating Currents

HKAL Physics Essay writing : Electromagnetism
Chapter 17 Alternating Currents
17.1 Energy dissipation in an a.c. circuit
I
V
IR
(a) In an a.c. circuit containing reactance  :
I=
V
  R2
2
, where V is the voltage across the circuit [V, I are r.m.s. values]
(b) Rate of heat dissipation, still, = I2R
17.2 RC Circuit
17.2.1
Definition of the reactance of a capacitor
It is a measure of the opposition of the capacitor to the passage of alternating current. The reactance
of a capacitor depends on its capacitance and the frequency of the applied a.c.
17.2.2
Derivation of the reactance of a capacitor
(a) First method
I = I0 sint =
I
0
dV
dQ
= C C
dt
dt

sin t dt = C dVC
VC = 
i.e. VC
1
1

cos t =
sin(t  )
C
C
2
has a phase retardation of /2 on current I.
Reactance = (VC)rms/(I)rms =
1
C
(b) Second method
a.c. source of voltage E = E0 sinωt, ω being the angular frequency and t the time
First of all neglect R, Current, I =
d(CE 0 sin t)
dQ
=
and I = CE0cos t
dt
dt
If R considered, C neglected. Current, I’ = E0/R sin t.
Thus for capacitor, effective impedance = I/C compared with R for the resistor.
Given condition implies that R < 1/C, i.e. p.d. across R can be neglected.
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17.2.3
Power delivered to the circuit
(a) For an a.c. source of voltage E = E0 sinωt, ω being the angular frequency and t the time,
current in circuit at any instant is given by equation I = CE0cos t.
While the power, P = IE = CE02 sint cost, i.e. P = ½CE02 sin2t
(b) Power flows out and into the a.c. source alternately at a frequency double that of the source.
17.3 LR Circuit
17.3.1
Derivation of the reactance of an inductor
(a) I = I0 sin t, VL = L
dI
= LI0 cos t or VL = LI0 sin (t + /2),
dt
i.e. VL has phase advance of /2 on current I
(b) Reactance = (VL)rms/(I)rms = L
17.3.2
Equation of state
back e.m.f.
p.d.
I
E
K
When closing key K
(a) back e.m.f., L
dI
.
dt
(b) p.d. across R, IR both act in opposite directions to e.m.f. E.
Applying Kirchoff Law to circuit, E - L
dI
- IR = 0.
dt
17.3.3
Solution for the current
(a) For an a.c. source of voltage E = E0 sinωt.
i.e. E0 sin t - L
dI
- IR = 0.
dt
(b) Assume the solution of I = P cos t + Q sin t, substitute it into the equation above, we obtain
E0 sin t - L(-P sin t + Q cos t) - R(P cos t + Q sin t) = 0
i.e. sin t (E0 + LP - RQ) + cos t (-LQ - RP) = 0
(c) This equation must be satisfied for all values of t.
(1) t = 0, LQ + RP = 0
(2) t = /(2), E0 + LP - RQ = 0
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(d) Solving for Q, equation (2) becomes Q(R + 2L2/R) = E0 and Q = E0 R /(R2 + 2L2)
Also P = - LQ/R = - E0 L/(R2 + 2L2)
E L
E R
cos t 
sin t
Expression for current, I = 
2
2 2
2
(R   L )
(R   2L 2 )
17.3.4
Energy dissipation as Joule heating
(e) Instantaneous power loss due to Joule heating = I2R, say Pdissipated.
Substituting, Pdissipated =
(f)
E R
( R   2 L 2 )2
2
(2L2cos2t - 2LR cos t sin t+ R2sin2t)
Considering the time average (over a complete cycle) term by term,
cos2t = ½(1 + cos 2t) = ½, cos t sin t = ½ sin 2t = 0, sin2t = ½(1 - cos 2t) = ½,
Thus Pdissipated =
E R
2( R 2   2 L 2 )
17.4 LC Oscillator
17.4.1
Definition of s.h.m.
(a) At all times the acceleration of the moving body is proportional to its displacement from a
fixed point and is directed towards this point.
(b) Daily Examples Swinging suspended ceiling lamp, boat bobbing up and down in the sea, child
on swing.
17.4.2
Derivation of the solution
C
V
(a) At any time, V = L
dI
Q
=
.
dt
C
But current, I = dQ/dt and
Hence
L
dI
d 2Q
=
.
dt
dt 2
Q
d 2Q
= 
.
2
LC
dt
(b) This equation has a solution Q = Q0 cos 0t, where 0 =
i.e. charge (and current) will oscillate with a frequency f0 =
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1
LC
1
2 LC
.
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17.4.3
Oscillating graph for Q and I with the capacitor fully charged first
Charge max
C
L
Charge max
(i)
(ii)
Current max
Charge max
(iii)
(iv)
Current max
Q

(v)
charge
I
time

E
M
energy:
electrical
(E ) only
M
magnetic
(M) only
E
 E
M
E
only
M
M
only
current
E
E
only
The p.d. across the capacitor C. VC = Q/C and so the voltage observed in C.R.O. will be of the
same wave form as that of Q.
(a) C, initially, has max. charge Q - ENERGY IN ELECTRICAL FIELD and I = 0.
Current I starts to flow due to p.d. across C and back e.m.f. set up across L
(greatest at first, and also dI/dt).
(b) Q = 0, all charge transferred from one plate of C to the other. I is in a maximum
(no back e.m.f. VC = Q/C) - ENERGY IN MAGNETIC FIELD.
(c) Q increases so that polarity of C (plates) is reversed I = 0 when back e.m.f.
VL = VC - ENERGY IN ELECTRICAL FIELD, Cycle continues.
17.4.4
Oscillating Energy
energy
Electric (in capacitor)
Magnetic (in inductor)
0
T/2
T time
Stored energies alternate between the capacitor and inductor.
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17.4.5 Comparison with mass-spring system
(a) Both involve two kinds of energy which are transferred between the two components within
the system, i.e. potential energy of the compressed or stretched spring and the kinetic energy of
the mass versus the electric energy in the capacitor and the magnetic energy in the inductor.
Mechanical
Electromagnetic
spring
1 2
kx
2
capacitor
mass
1
mv 2
2
inductor
v
dx
dt
i
1 Q2
2 C
1 2
LI
2
dQ
dt
d 2x  k
d 2Q  1
k
(

Q)  

x
)


LC
m
m
dt 2
dt 2
(b) Certain electromagnetic quantities “correspond” to certain mechanical ones
(
Q  x, i  v, C
17.4.6

1
, L  m.
k
Damping in LC oscillatory circuit
L
I
(a)
1
LC
R
C
Q
Total p.d. across circuit is given by :
Q
dI
d 2Q
dQ Q
L
 IR  0 or L 2  R
 0
C
dt
dt
dt C
(b) There is inevitable resistance somewhere in the circuit such as resistance in connecting wires,
contact resistance etc. (or Energy loss through the emission of EM waves due to the oscillation
of electrons.)
(c) The current oscillates with frequency close to f0 (with small resistance) but with decreasing
amplitude.
(d) Result
Current
Time
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17.5 Tuning Circuit
17.5.1
Working principle
When a radio circuit is tuned by making its natural frequency for electrical oscillations equal to
that of the incoming radio signal. The resonant frequency
depends on the values of the
1
capacitance and the inductance in the circuit (f 
), the smaller the capacitance/inductance
LC
is, the higher the resonant frequency results. The resistance (usually comes from the inductor) gives
rise to the damping.
17.5.2
Practical tuning circuit
Aerial
L, R
C
To receiver
Earth
(a) Radio signals from different transmitting stations induce e.m.f.s of various frequencies in the
aerial, which cause currents flowing in the aerial coil. Then currents of the same frequencies
are induced in coil L by mutual induction.
(b) If C is adjusted so that the resonant frequency of the LCR circuit equals to the frequency of the
wanted station, a large current and p.d. at that frequency only will develop across C.
(c) This selected and amplified p.d. is then applied to the next stage of the receiver.
(d) Improvement : Use an inductor of high L/R ratio so increases the resonant current and hence
the voltage across C.
17.6 LRC series circuit
17.6.1
Definition of forced vibration and resonance
(a) A dormant physical system A can be excited into forced oscillations by coupled energy transfer
from a sustained oscillating system B, though the induced oscillation amplitude will, in general,
be small.
(b) If the oscillation frequency of B is adjusted to the natural oscillation frequency of A, maximum
power transfer will occur and the amplitude of the induced oscillations in system A will
maximize – this phenomenon is called resonance.
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17.6.2
Impedance
V L - VC
= (L - 1/C )I
VL = LI
VR = IR
V
I
I
VC = I/C
or
Hence impedance, Z  V / I  R2  (L 
17.6.3
VR = IR
1 2
)
C
Experiment to measure the r.m.s. current
a.c. supply
C
L, R
a.c. supply
mA
L
C
R
mA
or
(a) Set up the above circuit, set the signal generator’s output to a value, say 3 V, with a measurable
current, and increase the frequency stepwise from a low value, say 10 Hz, check whether the
output is constant at the previous setting, 3 V.
(b) Then record the corresponding current readings on the a.c. milliammeter, when frequency
increases, the current reading rises and then drops.
17.6.4
Maximum energy stored in the capacitor and the inductor
C
L
R
V = V0 cos t
(a)
V02  I 02 [(L 
(VC)max
Imax
=
VL = LI
(L - 1 ) I
C
I
VC =
C
PHASOR
DIAGRAM
V
I
VR = IR
1 2
)  R2 ]
C

I0
C

V0
1

1 2

C
2 1/ 2
[(L 
) R ]
C
I0 =
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V0
1 2
[(L 
)  R 2 ]1/ 2
C
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(b) Hence
(EC)max


1
C(Vc )2max
2
V02
1
 2
1 2
2[(L 
)  R2 ]  C
C
1
2
(EL)max  LI02

LV02
1 2
2[(L 
)  R2 ]
C
(c) If (EC)max = (EL)max then we have
1
L
 2C
or  
1
LC
(d) Due to resistance in circuit, power supplied by the signal generator is needed to sustain
oscillations. All power generated is lost in this resistor.
17.7 Resonance, damping and phase relation
17.7.1
General description
(a) Oscillator frequency varied until a maximum current observed in the LCR circuit.
If the resistance R ~ 0, the Resonant frequency, f0 =
1
2 LC
, where L is the inductance and C
the capacitance.
(b) Damping of the induced current would reduce the sharpness of the frequency response and this
would be produced by increasing the resistance R.
17.7.2
Derivation of the relation
(a) Considering a capacitor C across which is applied V = V0 sin t.
Corresponding current, I =
dQ
d
(CV0 sin t ) = CV0 cos t, that is voltage across
=
dt
dt
capacitor lags behind current by /2.
Consider an inductor L through which is passed I = I0 sin t.
To keep this current flowing, the applied supplied voltage = back e.m.f. = L
L
dI
dt
=
d
(I 0 sin t ) = LI0 cos t, that the voltage across inductor is ahead of current by /2.
dt
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(b)
Coil
VL
VR
L
R
VC
C
V
For a pure resistance R, the current and voltage will be in phase. The coil will have both
resistance (R) and inductance (L). The instantaneous voltages across the components may be
represented by the following phasors :
VL = LI, /2 ahead of I; VR = IR, in phase with I, and VC =
1
I , /2 lag behind I.
C
V
VL-VC

I
VR
current
f
0
applied frequency
Phase
/2
f0
0
applied frequency
- /2
(c) From phasor diagram it can be seen that :
(1) At low frequencies VC >> VL and phase angle will be -ve, tend  -/2, that is I
ahead by /2.
(2) At high frequencies VL >> VC and  will be +ve, tend  +/2, that I lags applied voltage
by /2.
(3) At resonance, VL = VC (L 
1
) - at natural oscillation frequency of circuit. Then I = V/R
C
and is a maximum.
(4) Also at resonance  = 0, the current is in phase with the applied voltage.
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17.7.3
Dependence of the resonant current on the resistance
(a) For circuit with (I) smaller resistance and (II) larger resistance, the current-frequency graph is:
I
smaller
resistance (I)
larger
resistance (II)
f0
f
,
where f0 = resonant frequency.
(b) current I and voltage across resistor VR are in phase, I = VR/R
(c)
(d)
VL0
(e)

VR0
VR0
V0
V0
V0

VC0
VC0
For frequency < f0, V lags
behind VR, so VR0 < V0
current I0 =
17.7.4
VL0
VL0
VR0 V0

= Imax
R
R
VC0
For frequency = f0, V and VR
in phase, so VR0 = V0
 current I0 = Imax =
VR0
V0
R
For frequency > f0, V leads VR,
so VR0 < V0
current I0 =
VR0 V0

= Imax.
R
R
Practical example of resonance
(a) useful - L/C/R tuned circuit for radio, frequency trap for interfering signal, vibrating
galvanometer, musical instrument involving tensioned string (not organ pipe!) etc.
(b) non-useful - car-vibration at particular engine speed, unwanted oscillations in amplifiers at
high frequencies (stray C’s), oscillations in suspended bridges due to wind/marching across
etc.
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17.8 Application
17.8.l
Low-pass filter
(a) Resistance of resistor is independent of the applied frequency.
Impedance of inductor is directly proportional to the applied frequency.
Impedance of capacitor is inversely proportional to the applied frequency.
(b) When frequency is low, VC is large while VL is small and we have VC >> VR >> VL.
(c) So any one of the below circuits is a low-pass filter.
multifrequency
signal
17.8.2
Low f
Low f
Low f
LRC Smoothing Circuit
(a) Smoothing capacitor C1
L
+
rectified
unsmoothed
p.d.
-
+
C1
C2
d.c.
output
-
Initially C1 is charged up as the rectified input voltage increases. When the input voltage
decreases, charge and energy are released through the load.
As the time constant (or capacitance) is large, the p.d. across the capacitor cannot follow the
variations of the input voltage, thus a more or less constant p.d. is maintained.
charging
discharging
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(b) Time constant
charging
A
Rectified
unsmoothed
p.d.
C1
V1 =
d.c.
output
R
C
B
discharging
V1
Rectified
unsmoothed
input p.d.
Time
0
Time constant of charged capacitor is long compared with period of 1 cycle.
e.g. T
= (32  10-6)  (25  103) = 0.8 s > 1/50 = 0.02 s
17.8.3
Filter circuit
(a) Filter capacitor C2
L
+
rectified
unsmoothed
p.d.
-
(i)
+
C1
C2
d.c.
output
-
The ripples produced by C1 can be resolved into a steady d.c. component and an a.c.
component (ripples).
ripples from C 1
Steady d.c.
a.c.
(ii) When this voltage is applied across L and C2 in series. L offers a much greater impedance
to the a.c. component than C2. Hence most of the unwanted a.c. ripples appear across L.
(iii) For the d.c. component, C2 has infinite resistance, and the whole of this component is
developed across C2 except for the small drop due to the resistance of L.
(iv) The filter thus acts as a potential divider, separating d.c. from a.c. and giving steady d.c.
output across C2.
(b) Required inductance
+
C
Rectified 1
unsmoothed
p.d.
_
L
V1
C2
Ripple
p.d.
ripple greatly reduced
steady d.c.
output
1
2fC
L/C2 act as a potential divider at f = 50 Hz, e.g. C/ L =
2fL
Essay_notes_EM17_18
=
100
1
=
.
6000
60
12/15
Chapter 18 Electronics
18.1 Diode
18.1.1
Non-ohmic behaviour of a diode
I
V
0
(a) For a semi-conductor diode, the I-V graph shows that current passes when the p.d. is applied in
one direction but is almost zero when its acts in the opposite direction.
(b) A diode thus has a small resistance if the p.d. is applied one way round
resistance when the p.d. is reversed. It is thus non-ohmic.
18.1.2
but
a
very
large
Full-wave rectification
(a) Circuit
A
~
V
4
1
Rectified
output p.d.
2
3
B
(b) Result
p.d. across load
time
Alternating input p.d.
(c) Working principle
Rectifiers have low resistance to current flow in one direction and high resistance in opposite
direction. If A +ve rectifiers 1/2 conduct, on other half of cycle A -ve rectifiers 3/4 conduct.
Voltage output is uni-directional.
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18.2 Transistors
18.2.1
Input-output Characteristic
(a) Circuit to determine the input-output d,c, voltage characteristic of an NPN transistor operating
in the common emitter configuration.
+ 5.0 V
Rc
amplifier
input
Rb
Ri
100 
V
Vin
10 k
2 k
Ib
Ic
Vbe
output
Vout
V
Other components used are :
(1) Ri used as a potential divider to vary Vin.
(2) Rb used to limit current Ib so as to not exceed maximum allowable for transistor (increase
of Ib leads to an increasing p.d. across Rb reducing Vbe and Ib).
(3) Rc is a load resistor - variations in the collector current Ic produce corresponding
variations in the p.d. across Rc.
(b) Typical input-output voltage characteristic
V out/V
5
4
3
2
1
0 0.5 1
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2
3
4
5
V in/V
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18.2.2
Application of transistor
(a) Voltage amplification
Must use the linear (shaded part shown) of the characteristic with 0.5 V < Vin < 1.0 V – then
amplification is constant and no distortion of form of input signal. Need to bias transistor to
operate within these limits of characteristic.
(b) Switch
Clearly the output voltage Vout can be switched from 5.0 V to ~ 0 V by changing Vin from 0 V
to a value > 1 V. i.e. high 1 to low 0.
(c) NOR gate
NOR GATE
A
R1
input V1 R
2
+VCC (6 V)
Ic
RL
output Vo
Ib
A
B
B
output
NOR gate symbol
(two-input)
0
A
B
Output
0
0
1
1
0
0
0
1
0
1
1
0
(1) A or B disconnected (Vi ~ 0 V) output Vo high.
(2) Either A or B connected so that Vin > 1.0 V, say output Vo ~ 0 V
(d) OR gate
+VCC
OR
GATE
A
C
output
B
0
NOR GATE
NOT GATE
A
B
C
A
B
C
Output
0
0
1
0
1
0
0
1
0
1
0
1
1
1
0
1
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