# Chapter 9:ALTERNATING CURRENT AND VOLTAGE¶

## Example 9.1,Page number: 237¶

In [3]:
#Question:
"""Finding the angle at which the instantaneous value of voltage is 10 V."""

from math import asin,degrees

#Variable Declaration:
Vm=20.0               #Peak value of sinusoidal voltage(in Volts)

#Calculations:
v=10.0    #Instantaneous Voltage(in Volts)
angle=degrees(asin(v/Vm))

#Result:
print "(a)The angle at which the instantaneous value of voltage is 10V is %.2f degrees." %(angle)
print "(b)The maximum value of voltage is Vm=20 V.This occurs twice in one cycle at angles 90 degrees and 270 degrees."

(a)The angle at which the instantaneous value of voltage is 10V is 30.00 degrees.
(b)The maximum value of voltage is Vm=20 V.This occurs twice in one cycle at angles 90 degrees and 270 degrees.


## Example 9.2,Page number: 237¶

In [2]:
#Question:
"""Finding the time represented by a 60 degrees phase angle."""

#Variable Declaration:
ang_freq=2000.0       #Angular Frequency(in radians per second)

#Calculations:
f=ang_freq/(2*pi)
T=1/f
t=160e-06
t_60=(60.0/360)*T

#Result:
print "(a)The frequency is %.2f Hz." %(f)
print "(b)The angular frequency is %.2f rad/sec." %(ang_freq)
print "(c)The instantaneous voltage when t=60 micro seconds, is %e V." %(v)
print "(d)The time represented by 60 degrees phase angle is %e seconds." %(t_60)

(a)The frequency is 318.31 Hz.
(b)The angular frequency is 2000.00 rad/sec.
(c)The instantaneous voltage when t=60 micro seconds, is 3.917381e-02 V.
(d)The time represented by 60 degrees phase angle is 5.235988e-04 seconds.


## Example 9.3,Page number: 237¶

In [6]:
#Question:
"""Finding the equation for the instantaneous value of a sinusoidal voltage."""

#Variable Declaration:
V_peak_to_peak=20.0    #Peak-to-peak Voltage(in Volts)

#Calculations:
V_m=V_peak_to_peak/2
T=10.0e-03
f=1/T
ang_freq=2*pi*f
phi=180-degrees(asin(3.6/V_m))
t=12e-03

#Result:
print "(a)The equation of the given sinusoidal voltage is v(t)=%.2f sin(%.2ft+%.2f) V." %(V_m,ang_freq,phi)
print "(b)The value of voltage at 12ms would be %.3f V." %(v_12)

(a)The equation of the given sinusoidal voltage is v(t)=10.00 sin(628.32t+158.90) V.
(b)The value of voltage at 12ms would be -9.985 V.


## Example 9.4,Page number: 238¶

In [3]:
#Question:
"""Finding the equation for the instantaneous value of an alternating current."""

from math import pi,sin,asin

#Variable Declaration:
f=60.0                #Frequency of alternating current(in Hertz)
i_m=12.0              #Peak value of the alternating current(in Amperes)

#Calculations:
ang_freq=2*pi*f
t1=1.0/360
i_t1=i_m*sin(ang_freq*t1)
i2=9.6
t2=(asin(9.6/12))/ang_freq

#Result:
print "(a)The equation for the instantaneous value of alternating current is i= %d sin(%.2ft) A." %(i_m,ang_freq)
print "(b)The value of current at t=1/360 second is %.2f A." %(i_t1)
print "(c)The time taken to reach 9.6A for the first time is %e seconds." %(t2)

(a)The equation for the instantaneous value of alternating current is i= 12 sin(376.99t) A.
(b)The value of current at t=1/360 second is 10.39 A.
(c)The time taken to reach 9.6A for the first time is 2.459727e-03 seconds.


## Example 9.5,Page number: 244¶

In [4]:
#Question:
"""Finding the phase difference between two sinusoidal currents."""

from math import pi

#Calculations:
ang_freq=100.0*pi
f=ang_freq/(2.0*pi)
T=1/f
t=(30.0/360.0)*T

#Result:
print "The phase difference in terms of time is %e seconds." %(t)

The phase difference in terms of time is 1.666667e-03 seconds.


## Example 9.6,Page number: 248¶

In [6]:
#Question:
"""Finding the power consumed by a resistor."""

from cmath import phase

#Variable Declaration:
R=10                   #Resistance of resistance(in Ohms)
I=4+ 1j*3              #Alternating current phasor(in Amperes)

#Calculations:
I_mod=abs(I)
I_phase= phase(I)
I_rms=I_mod
P=pow(I_rms,2)*R

#Result:
print "The power consumed by the 10 ohm resistor is %.2f W." %(P)

The power consumed by the 10 ohm resistor is 250.00 W.


## Example 9.7,Page number: 250¶

In [5]:
#Question:
"""Finding the resultant current obtained by adding two alternating currents."""

from cmath import rect,phase

#Calculations:
I1=rect(10,0)
I=I1+I2
Im=abs(I)
I_phase=degrees(phase(I))

#Result:
print "The resultant current is %.2f A at a phase angle of %.2f degrees." %(Im,I_phase)

The resultant current is 26.46 A at a phase angle of 40.89 degrees.


## Example 9.8,Page number: 250¶

In [7]:
#Question:
"""Finding the rms value of the sum of two currents."""

from cmath import rect

#Calculations:
I1=rect((10*sqrt(2)),0)
I=I1+I2
Im=abs(I)
Irms=Im/(sqrt(2))

#Result:
print "The rms value of the sum of the currents is %.2f A." %(Irms)

The rms value of the sum of the currents is 26.46 A.


## Example 9.9,Page number: 250¶

In [8]:
#Question:
"""Finding the rms value of the resultant current."""

from cmath import rect,phase

#Calculations:
I1=rect(5,0)
I=I1+I2+I3
Im=abs(I)
Irms=Im/(sqrt(2))
I_phase=degrees(phase(I))

#Result:
print "The rms value of the resultant current that leaves the junction is %.2f A at a phase angle of %.2f degrees." %(Irms,I_phase)

The rms value of the resultant current that leaves the junction is 5.00 A at a phase angle of -15.00 degrees.


## Example 9.10,Page number: 251¶

In [9]:
#Question:
"""Finding the average and rms value of the resultant cuurent in a wire."""

from math import sqrt

#Variable Declaration:
I1rms=10.0            #Rms value of direct current(in Amperes)

#Calculations:
avg=10.0
I2rms=10.0/sqrt(2)
I_rms=sqrt(pow(I1rms,2)+pow(I2rms,2))

#Result:
print "The average value of the resultant current is %d A as the current goes as much positive as negative around the value of %d A." %(avg,avg)
print "The rms value of the resultant current is %.3f A." %(I_rms)

The average value of the resultant current is 10 A as the current goes as much positive as negative around the value of 10 A.
The rms value of the resultant current is 12.247 A.


## Example 9.11,Page number: 252¶

In [10]:
#Question:
"""Finding the average value of a voltage waveform."""

#Calculations:
area_0_to_1=10*(1e-03)
area_1_to_3=-5*(2e-03)
area_3_to_4=20*(1e-03)
area_4_to_5=0*(1e-03)
area_5_to_8=5*(3e-03)
total_area=(area_0_to_1+area_1_to_3+area_3_to_4+area_4_to_5+area_5_to_8)
total_period=8e-03
avg_value=total_area/total_period

#Result:
print "The average value of the alternating voltage waveform is %.3f V." %(avg_value)

The average value of the alternating voltage waveform is 4.375 V.


## Example 9.12,Page number: 252¶

In [11]:
#Question:
"""Finding the effective value of a voltage waveform."""

from math import sqrt

#Calculations:
area_0_to_10=400*(10e-03)
area_10_to_20=100*(10e-03)
total_area=(area_0_to_10+area_10_to_20)
total_period=20e-03
avg_value_of_square=total_area/total_period
rms=sqrt(avg_value_of_square)

#Result:
print "The effective(rms) value of the alternating voltage waveform is %.3f V." %(rms)

The effective(rms) value of the alternating voltage waveform is 15.811 V.


## Example 9.13,Page number: 253¶

In [12]:
#Question:
"""Finding the rms value,the average value and the form factor for a current waveform."""

from math import sqrt

#Variable Declaration:
period=3.0            #Time period of the current waveform(in seconds)

#Calculations:
Irms=sqrt( ((pow(10,2)*2)+(pow(0,2)*1))/3 )
Iavg=((10.0*2)+(0*1))/3.0
form_factor=Irms/Iavg

#Result:
print "The rms value of current waveform is %.2f A." %(Irms)
print "The average value of current waveform is %.2f A." %(Iavg)
print "The form factor of the current waveform is %.2f." %(form_factor)

The rms value of current waveform is 8.12 A.
The average value of current waveform is 6.67 A.
The form factor of the current waveform is 1.22.


## Example 9.14,Page number: 253¶

In [13]:
#Question:
"""Finding the form factor and the peak factor for a saw-tooth waveform."""

from math import sqrt

#Variable Declaration:
T=5e-03               #Time period of saw-tooth waveform(in seconds)
V_m=10.0              #Peak value of the saw-tooth voltage(in Volts)

""" Vav=(Area under the curve in one cycle)/(Duration of one cycle) """

#Calculations:
Vav=((1.0/2)*V_m*T)/T
Vrms=V_m/(sqrt(3))
form_factor=Vrms/Vav
peak_factor=V_m/Vrms

#Result:
print "The average value of the saw-tooth voltage waveform is %.3f V." %(Vav)
print "The rms value of the saw-tooth voltage waveform is %.3f V." %(Vrms)
print "The form factor for the saw-tooth voltage waveform is %.3f." %(form_factor)
print "The peak factor for the saw-tooth voltage waveform is %.3f." %(peak_factor)

The average value of the saw-tooth voltage waveform is 5.000 V.
The rms value of the saw-tooth voltage waveform is 5.774 V.
The form factor for the saw-tooth voltage waveform is 1.155.
The peak factor for the saw-tooth voltage waveform is 1.732.


## Example 9.15,Page number: 255¶

In [14]:
#Question:
"""Finding the average power,the apparent power,the instantaneous power and the power factor in percentage in an ac circuit."""

from math import cos,sqrt,pi

#Variable Declaration:
phase_angle=pi/5      #Phase difference between the alternating current and alternating voltage(in radians)
Vm=55                 #Peak value of the alternating voltage(in Volts)
Im=6.1                #Peak value of the alternating current(in Amperes)

#Calculations:
Vrms=Vm/sqrt(2)
Irms=Im/sqrt(2)
pf=cos(phase_angle)
P_avg=Vrms*Irms*cos(phase_angle)
P_app=Vrms*Irms
P_inst=P_avg-(Vrms*Irms*cos((2*0.3)-(pi/5)))

#Result:
print "The average power is %.2f W." %(P_avg)
print "The apparent power is %.2f VA." %(P_app)
print "The instantaneous power at wt=0.3 is %.2f W." %(P_inst)
print "The power factor is %.3f lagging." %(pf)

The average power is 135.71 W.
The apparent power is 167.75 VA.
The instantaneous power at wt=0.3 is -31.97 W.
The power factor is 0.809 lagging.


## Example 9.16,Page number:262¶

In [2]:
#Question:
"""Finding the average and rms values of waveforms."""

from math import sqrt

#Variable Declaration:
Vm1=10.0              #Peak voltage of first waveform(in Volts)
Vm2=10.0              #Peak voltage of second waveform(in Volts)

#Calculations:
Vav1=Vm1/2.0
Vrms1=Vm1/sqrt(3.0)
Vav2=Vm2/4.0
Vrms2=Vm2/sqrt(6.0)

#Result:
print "(a)The average value of the voltage is %.2f V and the rms value of the voltage is %.2f V." %(Vav1,Vrms1)
print "(b)The average value of the voltage is %.2f V and the rms value of the voltage is %.2f V." %(Vav2,Vrms2)

(a)The average value of the voltage is 5.00 V and the rms value of the voltage is 5.77 V.
(b)The average value of the voltage is 2.50 V and the rms value of the voltage is 4.08 V.


## Example 9.17,Page number:263¶

In [1]:
#Question:
"Finding the value of current at a given instant."""

from math import pi,sin,asin

#Variable Declaration:
Im=12.0               #Maximum value of the alternating current(in Amperes)
f=60.0                #Frequency of the alternating current(in Hertz)

#Calculations:
w=2*pi*f
t=1/360.0
i=Im*sin(w*t)
i1=9.6
t=asin(i1/Im)/w

#Result:
print "(a)The equation for the instantaneous current is i(t)=%.2f sin(%.2f*t) A." %(Im,w)
print "(b)The value of the current after (1/360) second is %.2f A." %round(i,2)
print "(c)The time taken to reach 9.6 A for the first time is %e seconds." %(t)

(a)The equation for the instantaneous current is i(t)=12.00 sin(376.99*t) A.
(b)The value of the current after (1/360) second is 10.39 A.
(c)The time taken to reach 9.6 A for the first time is 2.459727e-03 seconds.


## Example 9.18,Page number:263¶

In [16]:
#Question:
"""Finding the expression of an alternating current in cosine form."""

from math import asin,pi

#Variable Declaration:
Imax=10.0             #Maximum value of current(in Amperes)
Io=5.0                #Value of current at t=0(in Amperes)

#Calculations:
phi=asin(Io/Imax)
phi_new=(pi/2.0)-phi

#Result:
print "The expression for current is i=%.2f cos(wt-%.2f) A." %(Imax,phi_new)

The expression for current is i=10.00 cos(wt-1.05) A.


## Example 9.19,Page number:264¶

In [2]:
#Question:
"""Finding the time(from negative value) at which the instantaneous current is 10/sqrt(2.0) A."""

#Variable Declaration:
Irms=20.0             #Rms value of alternating current(in Amperes)
f=50.0                #Frequency of the alternating current(in Hertz)

#Calculations:
ang_freq=2*pi*f
Im=Irms*sqrt(2.0)
i=10.0*sqrt(2.0)
ph_lag=pi/2.0
t=(asin(i/Im)+ph_lag)/ang_freq

#Result:
print "The time(measured from negative value) at which instantaneous current will be 10/sqrt(2.0) is %.2f ms." %round((t*1000),2)

The time(measured from negative value) at which instantaneous current will be 10/sqrt(2.0) is 6.67 ms.


## Example 9.20,Page number:264¶

In [9]:
#Question:
"""Finding the average and rms value of an alternating current."""

from math import pow,sqrt

#Variable Declaration:
Idc=10.0              #Dc current(in Amperes)
Im=5.0                #Peak value of sinusoidal component(in Volts)

#Calculations:
Iav=Idc
Irms=sqrt((10*10)+pow((5.0/sqrt(2.0)),2))

#Result:
print "The average value of current is %.2f A." %(Iav)
print "The rms value of current is %.2f A." %(Irms)

The average value of current is 10.00 A.
The rms value of current is 10.61 A.


## Example 9.21,Page number:264¶

In [24]:
#Question:
"""Finding the sum of three alternating voltages."""

from cmath import rect,phase

#Calculations:
v_res=v1+v2+v3

#Result:
print "The resultant voltage is v=%.2f sin(wt+(%.2f degrees)) V." %(abs(v_res),degrees(phase(v_res)))

The resultant voltage is v=250.07 sin(wt+(90.01 degrees)) V.


## Example 9.22,Page number:265¶

In [19]:
#Question:
"""Finding the reactance offered by an inductor and a capacitor."""

from math import pi

#Variable Declaration:
L=0.2                 #Inductance of the inductor(in Henry)
C=10e-06              #Capacitance of the capacitor(in Farads)
f=100                 #Initial frequency of the ac input voltage(in Hertz)
f1=140                #New frequency of the ac input voltage(in Hertz)

#Calculations:
X_L=2*pi*f*L
X_C=1.0/(2*pi*f*C)
X_L1=2*pi*f1*L
X_C1=1.0/(2*pi*f1*C)

#Result:
print "(a)For a frequency of 100Hz, X_L=%.2f Ohms and X_C=%.2f Ohms." %(X_L,X_C)
print "(a)For a frequency of 140Hz, X_L=%.2f Ohms and X_C=%.2f Ohms." %(X_L1,X_C1)

(a)For a frequency of 100Hz, X_L=125.66 Ohms and X_C=159.15 Ohms.
(a)For a frequency of 140Hz, X_L=175.93 Ohms and X_C=113.68 Ohms.


## Example 9.23,Page number:265¶

In [6]:
#Question:
"""Finding the currents in each case."""

from cmath import rect,phase

#Calculations:
I1=rect(10,0)
I2=rect(10,0)
Ia=I1+I2
Ib=I1+I2
Ic=I1+I2
V=250.0+1j*0
X_L=1j*25.0
Id=V/X_L
X_C=-1j*25.0
I=5.0+1j*0
Ve=I*X_C

#Result:
print "(a)The unknown current is %.2f A at a phase angle of %.2f degrees." %(abs(Ia),degrees(phase(Ia)))
print "(a)The unknown current is %.2f A at a phase angle of %.2f degrees." %(abs(Ib),degrees(phase(Ib)))
print "(a)The unknown current is %.2f A at a phase angle of %.2f degrees." %(abs(Ic),degrees(phase(Ic)))
print "(a)The unknown current is %.2f A at a phase angle of %.2f degrees." %(abs(Id),degrees(phase(Id)))
print "(a)The unknown voltage is %.2f A at a phase angle of %.2f degrees." %(abs(Ve),degrees(phase(Ve)))

(a)The unknown current is 20.00 A at a phase angle of 0.00 degrees.
(a)The unknown current is 14.14 A at a phase angle of 45.00 degrees.
(a)The unknown current is 20.00 A at a phase angle of -90.00 degrees.
(a)The unknown current is 10.00 A at a phase angle of -90.00 degrees.
(a)The unknown voltage is 125.00 A at a phase angle of -90.00 degrees.