Lesson 28 (1) Driven RLC Circuit In a series RLC circuit with an AC power source, the circuit equation !! + !! + !! = ℇ! !"#$ gives rise to the phasor diagram in which !
! the current I is in common to all three elements, VR is in phase ! !
!
!
!
with I , VL leads I by 90° , and VC lags I also by 90° . The relation ℇ = !! +!! + !! is shown for the case VL 0 > VC 0 , which occurs when X L > XC Defining the impedance by 2
Z = R 2 + ( X L ! XC ) the rms current and the average power are given by ℇ!"#
!!"# =
!
ℇ!!"# !
! = !!"# ℇ!"# !"#$ =
1 !
!
! + !" − !"
As a function of frequency, the highest power occurs when 1
!" −
= 0 !"
!
The solution is !! = !" the frequency of the freely oscillating undamped circuit . This phenomeon is called resonance, and the frequency the resonant frequency. A plot of the average power as a function of freqenecy is shown. The width of the peak can be characterized by the frequencies when the power is half of the maximum power. This occurs when 1
!" −
= ±! !"
which can be written as 1 1
!
!! ∓ ! − !!! = 0 !ℎ!"! ! = !
!
The two positive solutions of these two equations are !! =
!!!
1
+
2!
!
1
+ !! =
2!
!!!
1
+
2!
!
−
1
2!
!
The quantity Δ! = !! − !! = ! is a measure of the width of the peak in the power curve. The Q-­‐factor is defined by !=
!! !! ! 1 !
=
=
Δ!
!
! !
and is a measure of the sharpness of the resonance. A high Q circuit in a receiver allows tuning to a broadcast channel without interference from channels of near-­‐by frequencies. Example: A tuning circuit consists of a resistor, a 2.0pF capacitor, and a variable inductor. It is used to receive FM radios, which has the frequency range from 88MHz to 108MHz. What should be the range of the inductance? What should the resistance be so that a 2% change in frequency results in the power becoming halved when tuned to any frequency in the range? Solution: The upper limit of inductance is for the lowest frequency in the FM range 1
1
L= 2 =
= 1.64 !10 "6 H = 1.64µ H 2
6
"12
! 0 C ( 2" ! 88 !10 ) ! 2.0 !10
The lower limit is for the highest frequency: 1
1
L= 2 =
= 1.09 !10 "6 H = 1.09µ H 2
! 0 C ( 2" !108 !10 6 ) ! 2.0 !10 "12
The required Q factor is 2 Q=
R=
!0
1
>
= 25 !! 2 " 0.02
1 L 1 Lmin
1 1.09 !10 "6
<
<
= 29.5# Q C 25 C
25 2 !10 "12
Example: In the circuit shown, the frequency is 60Hz, and the peak emf is 5.0V. At the time when the current through the resistor is 0.70A and rising, find the voltages across the resistor, the inductor, the capacitor. Also find the emf and show that it is equal to the sum of the voltages. Solution: First find the peak current. ! = 2" ! 60 = 377 rad / s 1
1
X L = ! L = 377 ! 0.01 = 3.77" XC =
=
= 1.33" !C 377 ! 0.002
2
Z = R 2 + ( X L ! XC ) = 32 + 2.44 2 = 3.87" ℇ!
5
!! =
=
= 1.29! !
3.87
Then find the peak voltages: VR0 = I 0 R = 1.29 ! 3 = 3.87" VL 0 = I 0 X L = 1.29 ! 3.77 = 4.86" VC 0 = I 0 XC = 1.29 !1.33 = 1.72"
Next find the phase of the current assuming I = I 0 cos! t 0.7
! t = cos!1
= ±57.1° 1.29
Because the current is increasing, ! t = !57.1° . This is also the phase of VR VR = VR0 cos! t = 3.87cos (!57.1° ) = 2.10V Since VL leads the current by 90° , VL = VL 0 cos! t = 4.86 cos (!57.1° + 90° ) = 4.86 cos32.9° = 4.08V Since VC lags the current by 90° , 3 VC = VC 0 cos! t = 1.72 cos (!57.1° ! 90° ) = 1.72 cos(!147.1° ) = !1.44V Thus, VR +VL +VC = 4.74V Finally find the phase of the emf relative to the current. It leads the current by X ! XC
2.44
! = tan !1 L
= tan !1
= 39.1° R
3
ℇ = ℇ! !"# !" + ! = 5!"# −57.1° + 39.1° = 5 cos −18° = 4.76! Thus, except for round-­‐off error, !! + !! + !! = ℇ (2) Transformer A transformer is a device to change AC voltages. It consists of two coils wound around a common iron core, in separate circuits know as the primary and secondary circuits. The primary coil is connected to an AC power source, and the secondary to a load. The voltages of both coils are related to the rate of change of magnetic flux, which is the same for each turn of the two coils because the magnetic field is guided by the iron core. Denoting this rate of magnetic flux change by d! m dt , then numbers of turns in the primary and secondary coils by N p and N s the voltages are given by Vp = !N p
d! m
dt
Vs = !N s
d! m
dt
It follows that the voltage is proportional to the number of turns: Vs N s
=
Vp N p
This relation applies also to the rms voltages. The voltage is stepped up by having more turns in the secondary than in the primary, and is stepped down vice versa. In an ideal transformer, the power delivered to the primary coils is completely transferred to the secondary coils. This results in the following relations between the currents in the circuits: I pVp = I sVs 4 so that the currents are inversely proportional to the number of turns: Is N p
=
I p Ns
In long distance power transmission, in order to reduce power loss from electrical resistance, the current is reduced by a step up transformer, and the voltage is stepped down before safely delivering to the households. Typically, the voltage coming out of the generator in the power station is 20kV, which is stepped up to 365kV in the transmission line, then stepped down to 4kV in the distribution line in the city, and finally stepped down to 240V for household use. (3) Complex numbers A complex number z takes the form z = x + iy where x, y are real numbers known as the real part and the imaginary part. It can be !
represented by the point ( x, y ) , or the vector xiˆ + yj , on the x-­‐y plane. The amplitude (magnitude )of the complex number is z = x 2 + y 2 and the argument is Addition and subtraction are defined by z1 ± z2 = ( x1 ± x2 ) + i ( y1 ± y2 ) y
! = tan !1 x
Geometrically, z1 + z2 is the vector sum of the vectors that represent z1 and z2 Multiplication satisfies distributive law, with i 2 = !1 : z1z2 = ( x1 + iy1 ) ( x2 + iy2 ) = x1 x2 + ix1 y2 + iy1 x2 + i 2 y1 y2 = x1 x2 ! y1 y2 + i ( x1 y2 + y1 x2 ) It follows that for any positive number a , i a i a = !a , so that we can take ( )( )
!a = ia ! !a = !ia as the two square roots of !a . The complex conjugate of z is z* = x ! iy 5 2
From the rule of multiplication zz* = x 2 + y 2 = z The reciprocal (multiplicative inverse) can therefore by defined by 1 z*
z1
1
and division by = 2 = z1 z z
z2
z2
A notation is defined for the complex number with unit magnitude and argument ! ei! = cos! + isin ! Using trigonometric identities cos!1 cos! 2 ! sin !1 sin ! 2 = cos (!1 + ! 2 ); sin !1 cos! 2 + cos!1 sin ! 2 = sin (!1 + ! 2 ) i ! +!
it follows that ei!1 ei!2 = e ( 1 2 ) By differentiating with respect to ! , d i!
e = !sin ! + i cos! = i ( cos! + isin ! ) = iei! d!
An indefinite integral is !e
i!
1
d! = ei! = "iei! i
Some special values are ei! 2 = i
ei! = !1 Any complex number can be written in the so called polar form y
z = Aei!
A = z = x 2 + y 2 ! = tan !1 x
i ! +" 2 )
Since iz = ei! 2 Aei! = Ae (
, multiplication of a complex number by i rotates the vector for that number by 90° n the complex plane. (4) RC circuit The use of complex numbers facilitates solution of linear differential equations with constant coefficients. For the differential equation for LC circuit 6 d 2Q
1
+ ! 2Q = 0 ! =
2
dt
LC
the solution Q (t ) is of course real. But if we solve the equation allowing Q to be complex, the real part of Q would still be a solution of the equation. This works only because the equation is linear. We try Q = ae! t where a and ! are complex constants. Substitution into the equation gives ! 2 = !" 2 which has the solution ! = ±i" . The constant a remains arbitrary, which we write in the polar form a = Aei! . Choosing ! = i" , we find the complex solution i ! t+!
Q = Ae ( ) The actual solution is found by taking the real part: Re (Q ) = A cos (! t + ! ) with arbitrary constants A and ! . The choice ! = !i" does not give a new solution. (5) RLC Circuit Here the differential equation is d 2Q 1 dQ
L
1 +
+ " 2Q = 0
!=
"=
2
dt
! dt
R
LC
Substituting the try solution Q = ae! t yields a quadratic equation for ! : 2
" 1 %
1
1
with the roots ! + ! + # 2 = 0 ! = ! ± $ ' ! # 2 # 2" &
"
2"
1
In the case > " , we have two real roots, giving rise to the two independent 2!
exponentially decaying solutions for the overdamped case. 2
7 In the underdamped case when 2
1
< " , we write 2!
2
! 1 $
! 1 $
!t 2 ! +i"1t
2
2
# & ' " = i " ' # & = i"1 so that the solution is Q = ae
" 2! %
" 2! %
Writing a = Aei! The actual solution is the real part Re (Q ) = Ae
!
t
2!
cos ("1t + ! ) (6) Complex Impedance The AC circuit equations are linear equations in voltages and currents, which are also linearly related. We can consider the voltages and currents to be complex in solving these equations and then take the real parts of the solutions at the end. Since the voltages and currents are sinusoidal functions, we imagine them to carry the time varying factors ei!t where ! is the AC frequency : V ~ ei!t I ~ ei!t The voltage-­‐current relation for the inductor dI
becomes VL = i! LI L = iX L I L V = L dt
and the voltage charge relation for a capacitor Q 1
1
becomes V = = ! I dt VC =
I C = !iXC I C C C
i!C
For a resistor, we still have VR = RI R For a driven RLC series circuit, the equation !! + !! + !! = ℇ! !"#$ is generalized to the complex equation !! + !! + !! = ℇ where the complex emf is ℇ = ℇ! ! !"# . Using the complex voltage-­‐current relations, we find 8 where ℇ
! = !
Z = R + i ( X L ! XC ) is the complex impedance. We next write 1 / Z in the polar form: R ! i ( X L ! XC ) R ! i ( X L ! XC )
1
1
1
=
= 2
=
e!i! 2
2
2
Z R + i ( X L ! XC ) R ! i ( X L ! XC ) R + ( X L ! XC )
R + ( X L ! XC )
where cos ! =
R
2
R + ( X L ! XC )
2
sin ! =
X L ! XC
2
R + ( X L ! XC )
2
The complex current in polar form is then ℇ!
!=
! ! !"!! !
!
! + !! − !!
The actual current is the real part: ℇ!
!"# =
!"# !" − ! !! + !! − !! !
The complex impedances can be used in more complicated circuits. 9