Chapter 11: RESONANCE IN AC CIRCUITS

Example 11.1,Page number: 315

In [7]:
#Question:
"""Finding the resonant frequency,quality factor,voltage across each element in a series RLC circuit."""

from math import sqrt,pi,pow

#Variable Declaration:
R=12.0                #Resistance of resistor(in Ohms)
L=0.15                #Self-inductance of inductor(in Henry)
C=100e-06             #Capacitance of capacitor(in Farads) 
V=100.0               #Voltage(rms) of ac source(in Volts)


#Calculations:
resonant_freq=1/(2.0*pi*sqrt(L*C))
I_max=V/R
freq_c=(sqrt((1.0/(L*C))-(0.5*pow((R/L),2))))/(2.0*pi)
freq_l=1.0/(sqrt((L*C)-(0.5*pow((R*C),2)))*2*pi)
cap_rea=1.0/(2.0*pi*freq_l*C)
ind_rea=2.0*pi*round(freq_l,2)*L
Q=cap_rea/R
Vr=V
Vl=Q*V
Vc=Vl


#Result:
print "(a)The Resonant frequency(at which the circuit current becomes maximum) is %.2f Hz." %(resonant_freq)
print "(b)The maximum current supplied by the source is %.2f A." %(I_max)
print "(c)The frequency at which voltage across the capacitor is maximum is %.2f Hz." %(freq_c)
print "(d)The frequency at which voltage across the inducttor is maximum is %.2f Hz." %(freq_l)
print "(e)The inductive reactance is %.2f Ohms." %(ind_rea)
print "(f)The capacitive reactance is %.2f Ohms." %(cap_rea)
print "(g)The quality factor of the circiut is %.2f." %(Q)
print "(h)The voltage drop across resistor is %.2f V." %(Vr)
print "   The voltage drop across inductor is %.2f V." %(Vl)
print "   The voltage drop across capacitor is %.2f V." %(Vc) 

(a)The Resonant frequency(at which the circuit current becomes maximum) is 41.09 Hz.
(b)The maximum current supplied by the source is 8.33 A.
(c)The frequency at which voltage across the capacitor is maximum is 40.10 Hz.
(d)The frequency at which voltage across the inducttor is maximum is 42.12 Hz.
(e)The inductive reactance is 39.70 Ohms.
(f)The capacitive reactance is 37.79 Ohms.
(g)The quality factor of the circiut is 3.15.
(h)The voltage drop across resistor is 100.00 V.
   The voltage drop across inductor is 314.91 V.
   The voltage drop across capacitor is 314.91 V.

Example 11.2,Page number: 316

In [4]:
#Question:
"""Finding the capacitance value to give resonance in a series RLC circuit. """

from math import pi,sqrt,pow

#Variable Declaration:
res_freq=50.0          #Resonant frequency(in Hertz)
L=0.5                  #Self-inductance of inductor(in Henry)
R=4.0                  #Resistance of resistor(in Ohms)
V=100.0                #Voltage of the supply(in Volts)


#Calculations:
C=1/(pow((2*pi*res_freq),2)*L)
I_max=V/R
V_L=I_max*(2*pi*res_freq*L)
V_C=V_L
Q=(2.0*pi*res_freq*L)/R


#Result:
print "(a)The capacitance to give resonance is %e F." %(C)
print "(b)The voltage across the inductor is %.2f V." %(V_L)
print "   The voltage across the capacitor is %.2f V." %(V_C)
print "(c)The quality factor of the circuit is %.2f." %(Q)

(a)The capacitance to give resonance is 2.026424e-05 F.
(b)The voltage across the inductor is 3926.99 V.
   The voltage across the capacitor is 3926.99 V.
(c)The quality factor of the circuit is 39.27.

Example 11.3,Page number: 317

In [5]:
#Question:
"""Finding the inductance,the circuit current and the voltage across the capacitor under resonance."""

from math import pi,pow

#Variable Declaration:
res_freq=175e03        #Resonant frequency(in Hertz) 
V=0.85                 #Voltage applied(in Volts)
Q=50.0                 #Quality factor of the coil
C=320e-012             #Capacitance of the capacitor(in Farads)


#Calculations:
L=1/(pow((2*pi*res_freq),2)*C)
ind_rea=2*pi*res_freq*L
R=ind_rea/Q
Io=V/R
Vc=Q*V


#Result:
print "The value of inductance is %e H." %(L)
print "The circuit current is %e A." %(Io)
print "The voltage across the capacitor under resonance is %.2f V." %(Vc)

The value of inductance is 2.584724e-03 H.
The circuit current is 1.495398e-02 A.
The voltage across the capacitor under resonance is 42.50 V.

Example 11.4,Page number: 317

In [6]:
#Question:
"""Finding the current at the resonant frequency and the energy stored by inductor."""

from math import pow,pi

#Variable Declaration:
res_freq=5e03          #Resonant frequency(in Hertz)
L=1e-03                #Self-inductance of the inductor(in Henry)
V=120.0                #Voltage of the supply(in Volts)
R=2.0                  #Resistance of the coil(in Ohms)


#Calculations:
C=1/(pow((2*pi*res_freq),2)*L)
I_max=V/R
""" U=0.5*L*I*I=L*Irms*Irms"""
U=L*I_max*I_max


#Result:
print "The required value of capacitance is %e F." %(C)
print "(a)The current at the resonance frequency is %.2f A." %(I_max)
print "(b)The maximum instantaneous energy is %.2f J." %(U)

The required value of capacitance is 1.013212e-06 F.
(a)The current at the resonance frequency is 60.00 A.
(b)The maximum instantaneous energy is 3.60 J.

Example 11.5,Page number: 318

In [7]:
#Question:
"""Finding the resonance frequency and the quality factor for the overall circuit."""

from math import sqrt,pi

#Variable Declaration:
R1=0.51                #Resistor of the resistor-1(in Ohms)  
R2=1.3                 #Resistor of the resistor-2(in Ohms) 
R3=0.24                #Resistor of the resistor-3(in Ohms)
L1=32e-03              #Self-inductance of the inductor-1(in Henry)
L2=15e-03              #Self-inductance of the inductor-2(in Henry)
C1=25e-06              #Capacitance of the capacitor-1(in Farads)
C2=62e-06              #Capacitance of the capacitor-2(in Farads)


#Calculations:
Req=R1+R2+R3
Leq=L1+L2
Ceq=(C1*C2)/(C1+C2)
res_freq=1/(2*pi*sqrt(Leq*Ceq))
Q=(sqrt(Leq/Ceq))/Req
Q1=(2*pi*res_freq*L1)/R1
Q2=(2*pi*res_freq*L2)/R2


#Result:
print "(a)The resonance frequency is %.2f Hz." %(res_freq)
print "(b)The quality factor of the overall circuit is %.2f." %(Q)
print "(c)The quality factor of coil-1 is %.2f." %(Q1)
print "(d)The quality factor of coil-2 is %.2f." %(Q2)

(a)The resonance frequency is 173.93 Hz.
(b)The quality factor of the overall circuit is 25.05.
(c)The quality factor of coil-1 is 68.57.
(d)The quality factor of coil-2 is 12.61.

Example 11.6,Page number: 320

In [8]:
#Question:
"""Finding the half-power frequencies of a series ac circuit."""

from math import pow,sqrt

#Variable Declaration:
bandwidth=75e03        #Bandwidth of the resonant circuit(in Hertz) 


#Calculations:
pro=pow((150e03),2)
sum=sqrt(pow(bandwidth,2)+(4*pro))
f2=(sum+bandwidth)/2.0
f1=(sum-bandwidth)/2.0


#Result:
print "Lower Half-power frequency is %e Hz." %(f1)
print "Upper Half-power frequency is %e Hz." %(f2)

Lower Half-power frequency is 1.171165e+05 Hz.
Upper Half-power frequency is 1.921165e+05 Hz.

Example 11.7,Page number: 324

In [11]:
#Question:
"""Finding the line current,quality factor and the dynamic impedance of a series-parallel ac circuit."""

from math import pi,sqrt,pow

#Variable Declaration:
L=200e-06              #Self-inductance of the inductor coil(in Henry)  
res_freq=1e06          #Resonant frequency(in Hertz)
R=20.0                 #Resistance of the coil(in Ohms)
Rs=8e03                #Series resistance(in Ohms)
V=230.0                #Voltage(rms) of the supply(in Volts) 


#Calculations:
C=1/(pow((2*pi*res_freq),2)*L)
XL=2*pi*res_freq*L
Q=XL/R
Zo=L/(C*R)
Z=Zo+Rs
I=V/Z


#Result:
print "(a)The value of capacitance to cause resonance is %e F." %(C)
print "(b)The Q factor of the circuit is %.5f." %(Q)
print "(c)The dynamic impedance of the parallel resonant circuit is %.2f Ohms." %(Zo)
print "(d)The total line current is %e A." %(I) 

(a)The value of capacitance to cause resonance is 1.266515e-10 F.
(b)The Q factor of the circuit is 62.83185.
(c)The dynamic impedance of the parallel resonant circuit is 78956.84 Ohms.
(d)The total line current is 2.644990e-03 A.

Example 11.8,Page number: 325

In [12]:
#Question:
"""Finding the resonant frequency,Q-factor and bandwidth of a practical parallel resonant circuit."""

from math import pow,sqrt,pi

#Variable Declaration:
R=150.0                #Resistance of the coil(in Ohms)
L=0.24                 #Self-inductance of the coil(in Henry)
C=3e-06                #Capacitance of the capacitor(in Farads)


#Calculations:
res_freq=(sqrt(1-((R*R*C)/L)))/(2*pi*sqrt(L*C))
Q=(2*pi*res_freq*L)/R
BW=res_freq/Q


#Result:
print "The resonant frequency is %.2f Hz." %(res_freq)
print "The quality factor is %.2f." %(Q)
print "The bandwidth is %.2f Hz." %(BW)

The resonant frequency is 159.02 Hz.
The quality factor is 1.60.
The bandwidth is 99.47 Hz.

Example 11.9,Page number: 326

In [16]:
#Question:
"""Finding the source frequency and the current supplied by the source."""

from math import sqrt,pi,pow,degrees
from cmath import phase

#Variable Declaration:
V=125.0               #Voltage of the source(in Volts)
C=20.5e-06            #Capacitance of the capacitor(in Farads)
R=1.06                #Resistance of the coil(in Ohms)
L=25.4e-03            #Inductance of the coil(in Henry)


#Calculations:
fo=1.0/(2*pi*sqrt(L*C))
Io=V/R
V_L=Io*(2*pi*fo*L)
V_C=V_L
X_L=(2*pi*fo*L)
Z_coil=R+(1j*X_L)
V_coil=Io*Z_coil
I=300.0/X_L
R_new=V/I
Rx=R_new-R


#Result:
print "(a) (i)The source frequency is %.2f Hz, and\n    (ii)The current supplied by the source is %.2f A.\n" %(fo,Io)
print "(b) (i)The voltage across the capacitor is %.2f V and" %(V_C)
print "    (ii)The voltage across the coil is %.2f V at an angle of %.2f degrees.\n" %(abs(V_coil),degrees(phase(V_coil)))
print "(c)The resistance that must be connected in series with the circuit to limit the capacitor voltage to 300V is %.3f Ohms." %(Rx)

(a) (i)The source frequency is 220.56 Hz, and
    (ii)The current supplied by the source is 117.92 A.

(b) (i)The voltage across the capacitor is 4150.92 V and
    (ii)The voltage across the coil is 4152.80 V at an angle of 88.28 degrees.

(c)The resistance that must be connected in series with the circuit to limit the capacitor voltage to 300V is 13.607 Ohms.

Example 11.10,Page number: 326

In [2]:
#Question:
"""Finding the maximum instantaneous energy stored in the inductor."""

from math import pow,pi,sqrt

#Variable Declaration:
R=3.0                 #Resistance of the coil(in Ohms)
L=12e-03              #Self-inductance of the coil(in Henry)
fo=9e03               #Resonant frequency(in Hertz)
V=240.0               #Supply voltage(in Volts)   


#Calculations:
C=1.0/(pow((2*pi*fo),2)*L)
Io=V/R
ener=0.5*L*Io*Io


#Result:
print "The value of capacitance to be connected in series with the coil is %e F." %(C)
print "The maximum instantaneous energy stored in the inductor is %.2f J." %(ener)

The value of capacitance to be connected in series with the coil is 2.605998e-08 F.
The maximum instantaneous energy stored in the inductor is 38.40 J.

Example 11.11,Page number: 327

In [3]:
#Question:
"""Finding the parameters of a series RLC circuit."""

from math import pi,sqrt,pow

#Variable Declaration:
fo=10e03              #Resonant frequency(in Hertz)
BW=1e03               #Bandwidth(in HErtz)
P=15.3                #Power drawn(in Watts)
V=200.0               #Voltage of generator(in Volts)


#Calculations:
V_R=V
R=(V_R*V_R)/P
""" Q=fo/BW=(2*pi*fo*L)/R; Q=Quality factor of the circuit. """
L=R/(2*pi*BW)
C=1.0/(pow((2*pi*fo),2)*L)


#Result:
print "The parameters of the circuit are:\n R=%.2f Ohms,\n L=%.3f H,\n C=%e F." %(R,L,C)

The parameters of the circuit are:
 R=2614.38 Ohms,
 L=0.416 H,
 C=6.087677e-10 F.

Example 11.12,Page number: 327

In [6]:
#Question:
"""Finding the half-power frequencies and the circuit current."""

from math import sqrt,pi,pow

#Variable Declaration:
fo=200.0              #Resonant frequency(in Hertz)
V=400.0               #Voltage of the source(in Volts)
R=20e-03              #Resistance of the coil(in Ohms)
L=6e-03               #Inductance of the coil(in Henry)


#Calculations:
C=1.0/(pow((2*pi*fo),2)*L)
Io=V/R
X_C=1.0/(2*pi*fo*C)
V_C=Io*X_C
Im=sqrt(2)*Io
U_max=0.5*L*Im*Im
Q=(2*pi*fo*L)/R
BW=fo/Q
f1=fo-(BW/2.0)
f2=fo+(BW/2.0)


#Result:
print "(a)The capacitance of the capacitor is %e F." %(C)
print "(b)The circuit current is %.2f kA." %(Io/1000)
print "(c)The voltage across the capacitor is %.2f kV." %(V_C/1000)
print "(d)The maximum energy stored in the coil is %.2f MJ." %(U_max/1000000)
print "(e)The lower half-power frequency is %.3f Hz and the upper half-power frequency is %.3f Hz." %(f1,f2)

(a)The capacitance of the capacitor is 1.055429e-04 F.
(b)The circuit current is 20.00 kA.
(c)The voltage across the capacitor is 150.80 kV.
(d)The maximum energy stored in the coil is 2.40 MJ.
(e)The lower half-power frequency is 199.735 Hz and the upper half-power frequency is 200.265 Hz.

Example 11.13,Page number: 327

In [17]:
#Question:
"""Finding the bandwidth,resonant frequency,inductance and capacitance."""

from math import pi,pow,sqrt

#Variable Declaration:
R=1e03                #Resistance of the resistor(in Ohms)
f1=20e03              #Lower half-power frequency(in Hertz) 
f2=100e03             #Upper half-power frequency(in Hertz)


#Calculations:
BW=f2-f1
res_freq=sqrt(f1*f2)
Q=res_freq/BW
L=(Q*R)/(2*pi*res_freq)
C=1.0/(pow((2*pi*res_freq),2)*L)


#Result:
print "(a)The bandwidth is %.2f kHz." %(BW/1000.0)
print "(b)The resonant frequency is %.2f kHz." %(res_freq/1000.0)
print "(c)The inductance is %e H." %(L)
print "(d)The capacitance is %e F." %(C)

(a)The bandwidth is 80.00 kHz.
(b)The resonant frequency is 44.72 kHz.
(c)The inductance is 1.989437e-03 H.
(d)The capacitance is 6.366198e-09 F.

Example 11.14,Page number: 328

In [14]:
#Question:
"""Finding the power at half-power frequencies."""

#Variable Declaration:
R=5.0                 #Resistance of resistor(in Ohms)
V=20.0                #Voltage of the source(in Volts)


#Calculations:
Zo=R
Io=V/Zo
Po=(Io*Io)*R
P_half=Po/2.0


#Result:
print "The power at half-power frequencies is %.2f W." %(P_half) 

The power at half-power frequencies is 40.00 W.

Example 11.15,Page number: 328

In [15]:
#Question:
"""Finding the half-power frequencies and the quality factor."""

from math import pi,pow,sqrt

#Variable Declaration:
res_freq=100.0        #Resonant frequency(in Hertz)
V=240.0               #Voltage of the source(in Volts)
R=55e-03              #Resistance of the coil(in Ohms)
L=7e-03               #Self-inductance of the coil(in Henry)


#Calculations:
C=1.0/(pow((2*pi*res_freq),2)*L)
Q=(2*pi*res_freq*L)/R
BW=res_freq/Q
f1=res_freq-(BW/2.0)
f2=res_freq+(BW/2.0)


#Result:
print "(a)The value of the capacitance is %e F." %(C)
print "(b)The quality factor of the circuit is %.2f." %(Q)
print "(c)The lower half-power frequency is %.2f Hz and The upper half-power frequency is %.2f Hz." %(f1,f2)

(a)The value of the capacitance is 3.618614e-04 F.
(b)The quality factor of the circuit is 79.97.
(c)The lower half-power frequency is 99.37 Hz and The upper half-power frequency is 100.63 Hz.

Example 11.16,Page number: 328

In [16]:
#Question:
"""Finding the resonance frequency and the effective resistance at resonance."""

from math import sqrt,pi

#Variable Declaration:
R=20.0                #Resistance of the coil(in Ohms)
L=0.2                 #Inductance of the coil(in Henry)
C=100e-06             #Capacitance of the capacitor(in Farads)


#Calculations:
res_freq=sqrt(1-((R*R*C)/L))/(2*pi*sqrt(L*C))
Zo=L/(C*R)


#Result:
print "The frequency at which the circuit behaves as a non-inductive reactance is %.2f Hz." %(res_freq)
print "The effective resistance at resonance is %.2f Ohms." %(Zo) 

The frequency at which the circuit behaves as a non-inductive reactance is 31.83 Hz.
The effective resistance at resonance is 100.00 Ohms.

Example 11.17,Page number: 328

In [18]:
#Question:
"""Finding the quality factor at the upper tuning frequency."""

from math import sqrt,pow,pi

#Variable Declaration:
L=20e-06              #Self-inductance of the coil(in Henry)
fo_1=570e03           #Lower tuning frequency(in Hertz) 
fo_2=1560e03          #Upper tuning frequency(in Hertz)
Q1=50.0               #Quality factor at the lower tuning frequency


#Calculations:
C1=1.0/(pow((2*pi*fo_1),2)*L)
C2=1.0/(pow((2*pi*fo_2),2)*L)
R=(2*pi*fo_1*L)/Q1
BW=fo_1/Q1
Q2=(2*pi*fo_2*L)/R


#Result:
print "(a)The range of tuning capacitor is from %.3f nF to %.3f nF." %((C2*1e09),(C1*1e09))
print "(b)The resistance of the coil is %.3f Ohms and the bandwidth of the circuit is %.3f kHz." %(R,(BW/1000))
print "(c)The quality factor of the circuit at the upper tuning frequency is %.3f." %(Q2) 

(a)The range of tuning capacitor is from 0.520 nF to 3.898 nF.
(b)The resistance of the coil is 1.433 Ohms and the bandwidth of the circuit is 11.400 kHz.
(c)The quality factor of the circuit at the upper tuning frequency is 136.842.