# Variables
V = 220.
I_a = 30. #armature currnet
R_a = 0.75 #Armature resistance
# Calculations
E_b = V - I_a*R_a # Since V = E_b+ I_a*R_a
# Results
print 'Induced EMF or back EMF in the motor is %.1f V'%(E_b)
# Variables
Pole = 4.
A = Pole #for lap winding
V = 230.
Z = 250. #number of armature conductors
phi = 30.*10**-3 #flux per pole in weber
I_a = 40.
R_a = 0.6 #Armature resistance
# Calculations
E_b = V - I_a*R_a # Since V = E_b+ I_a*R_a
N = E_b * 60*A/(phi*Pole*Z) #because E_b = phi*P*N*Z/(60*A)
# Results
print 'Back emf is %.0f V and running speed is %.0f rpm'%(E_b,N)
# Variables
Pole = 4.
A = Pole #for lap winding
Z = 480. #number of armature conductors
phi = 20.*10**-3 #flux per pole in weber
I_a = 50. #Armature current
# Calculations
T_a = 0.159*phi*I_a*Pole*Z/A #Gross torque developed by armature
# Results
print 'Gross torque developed by armature is %.3f N-m'%(T_a)
import math
# Variables
Pole = 4.
A = Pole #for lap winding
V = 230.
R_a = 0.8 #Armature resistance
N_0 = 1000. #no load speed in rpm
Z = 540. #number of armature conductors
phi = 25.*10**-3 #flux per pole in weber
# Calculations and Results
E_b0 = phi*Pole*N_0*Z/(60*A) #induced emf
#part(i)
print 'i)Induced e.m.f = %.0f V'%(E_b0)
#part(ii)
I_a0 = (V- E_b0)/R_a #because V = E_b0+ I_a0*R_a
print 'ii)Armature current = %.2f A'%(I_a0)
#part(iii)
stray_losses = E_b0*I_a0 #on no load ,power developed is fully power required to overcome strya losses
print 'iii)Stray loss = %.2f W'%(stray_losses)
#part(iv)
T_f = E_b0*I_a0/(2*math.pi*N_0/60) #lost torque
print 'iv)Lost torque = %.3f N-m'%(T_f)
# Variables
Pole = 4.
Z = 200. #No of armature conductors
A = 2. #wave connected armature
V = 250.
phi = 25.*10**-3 #flux per pole in weber
I_a = 60.
I_L = I_a #armature current
R_a = 0.15
R_se = 0.2 #resistances of armature and series field winding
# Calculations
E_b = V - I_a*(R_a+R_se) #induced emf
N = E_b * 60*A/(phi*Pole*Z) #because E_b = phi*P*N*Z/(60*A)
# Results
print 'Required speed is %.0f r.p.m'%(N)
# Variables
V = 250.
I_L = 20. #load current
R_a = 0.3
R_sh = 200. #Armature and shunt field winding
# Calculations
I_sh = V/R_sh #shunt current
I_a = I_L-I_sh #armature current
E_b = V - I_a*R_a #emf generated
# Results
print 'Armature current is %.2f A'%(I_a)
print 'Back e.m.f is %.3f V'%(E_b)
# Variables
V = 220.
R_a = 0.3
R_sh = 110. #resistance of armature and shunt field winding
#no load
N_0 = 1000. #no load speed in r.p.m
I_L0 = 6. #line current on no load
I_sh = V/R_sh #no load shnt current
I_a0 = I_L0 - I_sh #no load armature current
E_b0 = V - I_a0*R_a #no load induced emf
# Calculations
#full load
I_sh_FL = V/R_sh
I_L_FL = 50 #line current at full load
I_a_FL = I_L_FL - I_sh_FL #full load armature current
E_b_FL = V - I_a_FL * R_a #full load induced emf
#using speed equation as treating phi as constant
N_FL = N_0 * (E_b_FL/E_b0)
# Results
print 'Speed on full load is %.2f r.p.m'%(N_FL)
# Variables
R_a = 0.2
R_se = 0.3 #resistance of armature and series field winding
#following variables correspond to load 1
V = 250.
N_1 = 800.
I_1 = 20.
I_a1 = I_1
I_se1 = I_a1
# Calculations
E_b1 = V - I_a1*(R_a+R_se)
#following variables correspond to load 2
I_2 = 50.
I_a2 = I_2
E_b2 = V - I_a2*(R_a+R_se)
#from speed equation it can be derived that
N_2 = N_1 * (E_b2/E_b1) * (I_a1/I_a2)
# Results
print 'Speed on motor on no load is %.0f r.p.m'%(N_2)
# Variables
V = 250.
R_a = 0.3
R_sh = 200. #resistance of armature and shunt field winding
R_x = 150. #additional resistance added in series to field winding
I_L1 = 22.
I_sh1 = V/R_sh #initial shunt current before adding 150 ohms resistance
I_a1 = I_L1 - I_sh1 #initial armature current before adding 150 ohms resistance
N_1 = 1500. #initial speed before adding 150 ohms resistance
# Calculations
#T (prop.) phi*I_a (prop.) I_sh*I_a and T_1 = T_2 and simplifying further
I_sh2 = V/(R_sh + R_x) #new shunt current
I_a2 = I_sh1*I_a1/I_sh2 #New armature current
E_b1 = V - I_a1*R_a #induced emf before adding 150 ohms resistance
E_b2 = V - I_a2*R_a #new emf
N_2 = N_1 * (E_b2/E_b1) * (I_sh1/I_sh2) #new speed
# Results
print 'New armature current and speed are %.4f A and %.f r.p.m respectively'%(I_a2,N_2)
import math
# Variables
V = 250.
R_a = 0.15
R_se = 0.1
R_x = 0.1 #Resimath.tance of armature , series field winding and extra resistance
N_1 = 800. #initial speed before load torque is increased
I_1 = 30.
I_a1 = I_1
I_se1 = I_1 #initial currents
# Calculations
T_2_by_T_1 = 1 + (50./100) #50 percent increase as mentioned in question
I_se2_by_I_a2 = R_x/(R_x + R_se) #from the figure
#T (prop.) phi*I_a (prop.) I_sh*I_a and T_1 = T_2 and simplifying ,solving further
I_a2 = math.sqrt(I_a1*I_se1*T_2_by_T_1/I_se2_by_I_a2) #new armature current
I_se2 = I_se2_by_I_a2 *I_a2 #new series field current
E_b1 = V - I_a1*R_a - I_se1*R_se #indiced emf initially
E_b2 = V - I_a2*R_a - I_se2*R_se #new induced emf
N_2 = N_1 * (E_b2/E_b1) * (I_se1/I_se2) #required speed
# Results
print 'The required running speed of motor is %.3f r.p.m'%(N_2)
import math
# Variables
V = 220.
I_1 = 50.
I_a1 = I_1 #Currents before adding extra resistance
T_2_by_T_1 = 0.5
R_t = 0.15 #R_e + R_se = 0.15
# Calculations
I_a2 = I_a1 * math.sqrt(T_2_by_T_1) #Because T (prop.) I_a**2
E_b1 = V-I_a1*(R_t) #induced emf before adding extra resistance
N_1 = 500.
N_2 = 300. #speeds before and adding extra resistance
#N (prop.) E_b/phi (prop.) E_b/I_a
E_b2 = E_b1 *(I_a2/I_a1)*(N_2/N_1) #induced emf after adding resistance
R_x = (V-E_b2)/I_a2 -R_t #because E_b2 = V - I_a2*(R_a + R_se + R_x)
# Results
print 'Desired extrea resistance = %.4f ohms '%(R_x)
import math
# Variables
R_a = 1.
I_a = 1.2
V = 50.
# Calculations and Results
#part(i)
E_b = V - I_a*R_a
rot_loss_NL = E_b*I_a #no load rotational loss
print 'i)No load rotational losses = %.2f W'%(rot_loss_NL)
#part(ii)
omega_2000 = 2*math.pi*2000/60 #angular velocity when speed of motor = 2000 rpm
K_m = E_b/omega_2000 #to determine K_m
V = 48.
omega_1800 = 2*math.pi*1800/60 #angular velocity when speed of motor = 1800 rpm
E_b = K_m*omega_1800
I_a = (V-E_b)/R_a #armature current
P_dev = E_b*I_a #power developed
motor_output = P_dev - rot_loss_NL
print 'ii)Motor output = %.f W'%(motor_output)
#part(iii)
E_b = 0. #when motor stalls
V_stall = 20. #voltage during stalling
I_a = V_stall/R_a #armature current during stalling
T_stall = K_m*I_a #stalling torque
print 'iii)Stalling torque = %.2f N-m'%(T_stall)
print 'partii answer is slightly different due to inaccurate calculation of Power developed'
# Variables
V = 120.
R_a = 0.2
R_sh = 60. #armature and field resistance
I_L1 = 40.
N_1 = 1800.
# Calculations
I_sh = V/R_sh
I_a1 = I_L1 - I_sh
E_b1 = V -I_a1*R_a #Induced emf at half load
T2_by_T1 = 1./2
I_a2 = I_a1*(T2_by_T1) #T (prop.)I_a
E_b2 = V- I_a2*R_a #induced emf at half load
N_2 = N_1 *(E_b2/E_b1) #N (prop.) E_b as phi is constant
# Results
print 'Speed on half load condition is %.2f r.p.m'%(N_2)
# Variables
R_a = 0.08
E_b1 = 242.
V = 250.
# Calculations and Results
#part(i)
I_a1 = (V-E_b1)/R_a
print 'i)Armature current = %.0f A'%(I_a1)
#part(ii)
N = 0.
E_b = 0. #because N = 0
I_a_start = V/R_a
print 'ii)Starting armature current = %.0f A'%(I_a_start)
#part(iii)
I_a2 = 120.
E_b2 = V-I_a2*R_a
print 'iii)Back emf if armature current is changed to 120 A = %.1f V'%(E_b2)
#part(iv)
I_a = 87.
N_m = 1500.
E_g = V + I_a*R_a #induced emf
N_g = N_m*(E_g/E_b1) #as E (prop.) N
print 'iv)Generator speed to deliver 87 A at 250 V = %.1f rpm'%(N_g)
import math
from numpy import roots
# Variables
shaft_output = 80.*746 #coverted to watts
eta = 80./100 #efficiency
V = 250.
N_1 = 1200.
R_a = 0.04
R_sh = 250. #armature and shunt field resistance
# Calculations and Results
power_input = shaft_output/eta
I_L = power_input /V
I_sh = V / R_sh
I_a = I_L - I_sh
E_b1 = V - I_a*R_a
gross_mechanical_power = E_b1*I_a #electrical equivalent of mechanical power developed
stray_losses = gross_mechanical_power - shaft_output
print 'Mechanical power developed on full load = %.3f kW'%(gross_mechanical_power/1000)
#on no load shaft_output = 0 and entire gross power is used to overcome stray losses
Eb0_Ia0 = stray_losses
#E_b0 = V - I_a0*R_a ... solving for I_0
p = [R_a, -V, Eb0_Ia0]
ans = roots(p)
I_a0 = ans[1] #first root is ignored math.since its too large
I_L0 = I_sh+I_a0 #current drawn from supply
E_b0 = V - I_a0*R_a
#From speed equation N (prop.) E_b
N_0 = N_1*(E_b0/E_b1)
print 'No load speed and current are %.4f rpm and %.2f A respectively'%(N_0,I_L0)
# Variables
V = 250.
P = 4.
R_a = 0.1
R_sh = 124. #armature and shunt field resistance
I_L0 = 4.
N_0 = 1200.
I_L_1 = 61.
# Calculations
I_sh = V/R_sh
I_a0 = I_L0-I_sh
V_brush = 2 #voltage loss due to brush
E_b0 = V - I_a0*R_a- V_brush
I_a1 = I_L_1 - I_sh
E_b1 = V - I_a1*R_a -V_brush
phi1_by_phi0 = 1-(5./100) #weakened by 5 %
N_1 = N_0 *(E_b1/E_b0) /phi1_by_phi0
# Results
print 'Full load speed is %.3f r.p.m'%(N_1)
# Variables
V = 250.
R_a = 0.15
R_sh = 167.67 #armature and shunt field resistance
N_0 = 1280. #speed at no load
#full load
I_L1 = 67. #current drawn on full load
I_sh = V / R_sh #as shunt motor
I_a1 = I_L1- I_sh
E_b1 = V - I_a1*R_a
#on no load
I_L0 = 6.5
I_a0 = I_L0 - I_sh
E_b0 = V - I_a0*R_a
# Calculations and Results
#part(i) USING SPEED EQUATION
#N (prop.) E_b/phi (prop.)E_b #as phi is constant
N_1 = N_0 * (E_b1 / E_b0)
print 'i)Full load speed = %.3f r.p.m'%(N_1)
#part(ii)
speed_regulation = 100* ((N_0-N_1)/N_1)
#N_1 is full load speed and N_0 = No load speed
print 'ii)Speed regulation = %.2f percent '%(speed_regulation )
#part(iii)
shaft_output_FL = E_b1*I_a1 - E_b0*I_a0 #full load power developed - stray losses
hp_rating = shaft_output_FL /746
print 'iii)HP rating of machine = %.2f h.p'%(hp_rating)
#part(iv)
power_input = V*I_L1
eta = 100*(shaft_output_FL/power_input) #full load efficiency
print 'iv)Full load efficiency = %.2f percent'%(eta)
# Variables
V = 200.
R_a = 0.5
R_se = 0.2
R_x = 0.2 #armature and series field resistance; extra resistance
I_a1 = 20.
I_1 = I_a1
I_se1 = I_a1
I_a2 = 20.
I_2 = I_a2
I_se2 = I_2 *(R_x/(R_se+R_x))
# Calculations
E_b1 = V -I_a1*R_a - I_a1*R_se
E_b2 = V -I_a2*R_a - I_se2*R_se
phi2_by_phi1 = 70./100
N_1 = 1000
N_2 = N_1*(E_b2/E_b1) /phi2_by_phi1 #N (prop.) E_b/phi
# Results
print 'Required speed is %.2f r.p.m'%(N_2)
import math
# Variables
V = 110.
P = 4.
R_a = 0.1
R = 0.01 #A resistance of 0.01 ohms
R_se = R+R
# Calculations
#case(i)
I_1 = 50.
I_a1 = I_1
N_1 = 700.
E_b1 = V -I_a1*(R_a + R_se)
#T (prop) phi*I_a from torque equation (1)
#phi_1 (prop.) I_a1 (2)
#case(ii) when I_a2 gets divided to half
#phi_2 (prop.) I_a2/2 (3)
#combining (1)(2)(3) and T1 = T2
I_a2 = math.sqrt(2*I_a1**2)
R_se_eqvt = (R*R)/(R+R) #Equavalent of parallel combination
E_b2 = V - I_a2*R_a - I_a2* R_se_eqvt
#using speed equation N (prop.) E_b / phi and using (2) and (3)
N_2 = N_1 *( E_b2/E_b1) *(I_a1/(I_a2/2))
# Results
print 'Speed after reconnection = %.3f r.p.m'%(N_2)
import math
# Variables
P = 4.
I_a1 = 50.
N_1 = 2000.
V = 230.
# Calculations and Results
#phi_1 is proportioanl to total ampere-turns produced by field coils
#phi_1 (prop.) I_a1*P*n (prop.) 200*n (1)
#After reconnection ,phi_2 proportional to ampere turns divided as follows
#phi_2 (prop.) [I_a2/2*2*n + I_a2/2*2*n] (prop.) 2*n*I_a2 (2)
# Dividing (1) and (2) ,(phi_1/phi_2) = 100 / I_a2 (3)
#T (prop.) phi*I_a AND T (prop.) N**2 (4)(5)
#therefore N**2 (prop.) phi*I_a (6)
#N (prop.) E_b/phi (prop.) 1/phi ..
#Because drops across windings can be neglected , E_b1 = E_b2
#therefore N (prop.) 1/phi (7)
#using (7) and (6) phi**3 (prop.) 1/I_a (8)
#combining (3) and (8)
I_a2 = (50.*100**3)**(1./4) #new armature current
print 'New armature current = %.3f A'%(I_a2)
#combining (6) and (7) N**3 (prop.) I_a1
N_2 = N_1 *(I_a2/I_a1)**(1./3)
print 'New motor speed = %.3f r.p.m'%(N_2)
import math
# Variables
V = 200.
I_a1 = 30.
R_t = 1.5 #R_a + R_se
E_b1 = V - I_a1*R_t
N2_by_N1 = (60./100)
# Calculations
#T (prop.) I_a**2 and T (prop.) N_3....therefore I_a**2 (prop.) N**3
I_a2 = I_a1*math.sqrt(N2_by_N1**3)
#N (prop.) E_b/I_a
N2_by_N1
E_b2 = E_b1 *(I_a2/I_a1)*N2_by_N1
R_x = (V- E_b2)/I_a2 - R_t #because E_b2 = V - I_a2*(R_x+R_t)
# Results
print 'Additional resistance to be added in series with motor circuit = %.3f ohms'%(R_x)
import math
from numpy import array,roots
# Variables
V = 250.
R_a = 0.4
R_sh = 100. #armature and shunt field resistance
I_sh1 = V / R_sh
P_out_FL = 10 * 735.5
eta = 85./100 #efficiency
P_in = P_out_FL/eta
I_L1 = P_in /V
I_a1 = I_L1 - I_sh1
# T (prop.) phi*I_a (prop.) I_sh*I_a because phi (prop.) I_sh
#Bu torque is constant..
Ia2_Ish2 = I_a1*I_sh1
E_b1 = V - I_a1*R_a
#N (prop.) E_b/I_sh
#put E_b2 = V - I_a2*R_a and solving further for I_sh2 we get ,I_sh2**2 - 1.8824 I_sh2 +0.2417 = 0
p = array([1, -1.8824, 0.2417])
ans = roots(p)
I_sh2 = ans[0]
#root 1 was considered because its always easier to attain root(1) because less resistacne is needeed
#R_x in series with field
R_x = (V/I_sh2) -R_sh #because I_sh2 = V/(R_sh + R_x)
print 'Extra resistance to be added = %.2f ohms'%(R_x)
# Variables
R_t = 1. #R_t = R_se + R_a
V_1 = 230.
N_1 = 300.
N_2 = 375.
I_1 = 15.
I_a1 = I_1
# Calculations
#T (prop.) I_a**2 and T (prop.) N_2....therefore I_a**2 (prop.) N**2
I_a2 = I_a1 *(N_2/N_1)
E_b1 = V_1 - I_a1*(R_t)
#N (prop.) E_b/I_a
E_b2 = E_b1*(I_a2/I_a1)*(N_2/N_1)
V_2 = E_b2 + I_a2* (R_t) #because E_b2 = V_2 - I_a2*(R_a+R_se)
# Results
print 'Voltage supply needed = %.4f V'%(V_2)
# Variables
I_L1 = 30.
V = 230.
R_sh = 230.
R_a = 1.
I_sh = V / R_sh
I_a1 = I_L1 - I_sh
E_b1 = V - I_a1*R_a
# Calculations
#T (prop.) phi*I_a (prop.) I_a as phi is constant
#and torque is constant
I_a2 = I_a1
N2_by_N1 = 1./2
#N (prop.) E_b/phi (prop.) E_b
E_b2 = E_b1 *(N2_by_N1)
R_x = (V- E_b2)/I_a2 - R_a #Because E_b2 = V - I_a2*(R_a + R_x)
# Results
print 'resistance to be inserted in series = %.4f ohms '%(R_x)
# Variables
T_1 = 40. #initial torque
#phi_1 is initial flux
#phi_2 is new flux
#T_2 is new torque
#I_a1 is initial current
#I_a2 is new current
phi2_by_phi1 = 1- (30./100) #decrease by 30 percent
Ia2_by_Ia1 = 1+(15./100) #increase by 15 percent
# Calculations
#T (prop.)phi*I_a
T_2 = T_1*(phi2_by_phi1)*(Ia2_by_Ia1)
# Results
print 'New torque is %.1f N-m'%(T_2)
# Variables
V = 230.
N_1 = 1000.
N_2 = 950.
R_a = 0.5
R_sh = 230. #armature and shunt field resistance
I_L1 = 10.
# Calculations
I_sh = V/R_sh
I_a1 = I_L1 - I_sh
#T (prop.) phi*I_a (prop.) I_a with phi constant and T is constant due to full-load
I_a2 = I_a1
E_b1 = V - I_a1*R_a
E_b2 = E_b1*(N_2/N_1) #N (prop.) E_b /phi (prop.) E_b as phi is constant
R_x = (V-E_b2)/I_a2 -R_a
# Results
print 'resistance to be inserted in series with armature = %.4f ohms'%(R_x)
# Variables
V = 250.
N_0 = 1000.
I_0 = 5.
R_a = 0.2
R_sh = 250. #armature and shunt field resistance
I_L = 50. #on no load
I_sh = V / R_sh
I_a0 = I_0 - I_sh
I_a = I_L - I_sh
E_b0 = V- I_a0*R_a
E_b1 = V- I_a *R_a
# Calculations
phi1_by_phi0 = 1-(3./100) #weakens by 3 percent
#N (prop.) E_b/phi
N_1 = N_0 *(E_b1/E_b0) /phi1_by_phi0
# Results
print 'Speed when loaded and drawing 50A current is %.3f r.p.m'%(N_1)
import math
# Variables
V = 230.
I_a0 = 3.3
R_a = 0.3
R_sh = 160. #armature and shunt field resistance
I_L1 = 40.
N_0 = 1000.
E_b0 = V - I_a0*R_a
I_sh = V/ R_sh
I_a1 = I_L1 - I_sh
E_b1 = V - I_a1*R_a
phi1_by_phi0 = 1- (4./100) #weakening by 4 percent
# Calculations and Results
N_1 = N_0 *(E_b1/E_b0)/(phi1_by_phi0) #because N (prop.) E_b/phi
print 'Full load speed is %.4f rpm'%(N_1)
T_0 = E_b0*I_a0/(2*math.pi*N_0/60)
T_1 = T_0*(I_a1/I_a0)*phi1_by_phi0 # because T (prop.) phi*I_a
print 'Full load developed torque is %.4f N-m'%(T_1)
# Variables
V = 220.
I_L = 52.
N_1 = 750.
N_2 = 600.
R_a = 0.2
R_sh = 110. #armature and shunt field resistance
# Calculations and Results
I_sh = V/ R_sh
I_a1 = I_L - I_sh
I_a2 = I_a1 #T (prop.) I_a and T is constant
E_b1 = V - I_a1*R_a
#N (prop.) E_b/phi (prop.) E_b
E_b2 = E_b1*(N_2/N_1)
R_x = (V- E_b2)/I_a2 -R_a #Because E_b2 = V - I_a2*(R_a+R_x)
print 'resistance to be connected in series = %.2f ohms'%(R_x)
#After R_x gets connected in series with armature and 110 ohms in series with field winding
N_1 = 600.
I_sh2 = V /(R_sh+110)
I_a1 = 50.
I_sh1 = 2.
I_sh2 = 1.
#T (prop.) I_a*I_sh and T doesn't vary
I_a2 = I_a1*(I_sh1/I_sh2)
E_b1 = V - I_a1*(R_a+R_x)
E_b2 = V - I_a2*(R_a+R_x)
N_2 = N_1*(E_b2/E_b1)*(I_sh1/I_sh2) #Because N (prop.) E_b/I_sh
print 'New speed = %.3f rpm'%(N_2)
# Variables
V = 230.
R_a = 0.15
R_sh = 250. #armature and shunt field resistance
I_a1 = 50.
I_a2 = 80.
N_1 = 800.
N_2 = 1000.
I_sh1 = V / R_sh
# Calculations
E_b1 = V - I_a1*R_a
E_b2 = V - I_a2*R_a
I_sh2 = I_sh1*(E_b2/E_b1)*(N_1/N_2) #Because N (prop.) E_b/ I_sh
R_x = (V/I_sh2 ) - R_sh #because I_sh2 = V /(R_x+ R_sh)
# Results
print 'resistance to be added is R_x = %.0f ohms'%(R_x)
# Variables
V = 230.
R_a = 0.5
N_1 = 800.
N_2 = 600.
I_a2 = 20.
I_a1 = I_a2
E_b1 = V - I_a1*R_a
# Calculations
#N (prop.) E_b/phi (prop.) E_b as phi is constant
E_b2 = E_b1 *(N_2/N_1)
#additional resistance required
R_x = (V -E_b2)/I_a2 - R_a #because E_b2 = V - I_a2*(R_a+R_x)
# Results
print 'Additional resistance required = %.2f ohms '%(R_x)
# Variables
V = 220.
R_a = 0.5
R_x = 5. #armature resistacne and extra resistance
I_1 = 15.
I_se1 = I_1
I_se2 = I_se1
I_2 = I_se2
N_1 = 800.
# Calculations
E_b1 = V - I_1*R_a
E_b2 = V - I_2*(R_a+R_x)
N_2 = N_1*(E_b2/E_b1)*(I_se1/I_se2) #because N (prop.) E_b/I_se
# Results
print 'New speed of rotor = %.3f r.p.m'%(N_2)
# Variables
V = 250.
I_a1 = 20.
R_a = 0.5
N_1 = 1000.
N_2 = 500.
# Calculations and Results
#T (prop.) I_a and T_1 = T_2
I_a2 = I_a1
E_b1 = V - I_a1*R_a
#N (prop.) E_b
E_b2 = E_b1 *(N_2/N_1)
R_x = (V-E_b2)/I_a2 - R_a #because E_b2 = V - I_a2*(R_a+R_x)
print 'Additional resistance = %.0f ohms'%(R_x)
T3_by_T2 = 0.5 #torque is halved
I_a3 = I_a2 *(T3_by_T2) #new armature current
E_b3 = V - I_a3*(R_x + R_a)
N_3 = E_b3*N_2 / E_b2 #N (prop.) E_b
print 'New speed = %.3f rpm'%(N_3)
# Variables
P_out = 100*735.5
V = 500.
P = 4.
A = 2. # due to wave winding
Z = 492. #no of conductors
phi = 50.*10**-3 #flux per pole
eta = 92./100 #efficiency
P_in = P_out/eta
R_a = 0.1
R_sh = 250. #amature and shunt field resistance
# Calculations
I_L = P_in/V
I_sh = V/ R_sh
I_a = I_L - I_sh
E_b = V - I_a*R_a
N = E_b*60*A/(phi*P*Z) #because E_b = phi*P*N*Z/(60*A)
T_sh = P_out/(2*math.pi*N/60) #Useful torque
# Results
print 'i)Speed at full load = %.4f rpm'%(N)
print 'ii)Useful torque = %.2f N-m'%(T_sh)
print 'Answer mismatches due to improper approximation'
from numpy import roots
import math
# Variables
N_1 = 1000.
I_1 = 50.
I_a1 = I_1
V = 250.
R_x = 4.4
R_t = 0.6 #R_t = R_a+R_se
E_b1 = V - I_a1*(R_t)
# Calculations
#T (prop.)I_a**2 T (prop.) N**2 .... hence N (prop.) I_a
#N (prop.) E_b /I_a
#combining both E_b (prop.) I_a**2
#using E_b2 = V - I_a2*(R_a + R_se + R_x) and solving for I_a2 we get 0.088 I_a2**2 +5 I_a2 -250 = 0
p = [0.088 ,5, -250]
ans = roots(p)
I_a2 = ans[1] #root(1) is ignored as it is -ve
E_b2 = V - I_a2*(R_t + R_x)
N_2 = N_1*(E_b2/E_b1)*(I_a1/I_a2)
# Results
print 'Motor speed = %.2f r.p.m'%(N_2)
from numpy import roots
import math
# Variables
V = 250.
I_a1 = 20.
R_sh = 250.
R_a = 0.5 #shunt field and armature resistance
I_sh1 = V / R_sh
E_b1 = V - I_a1*R_a
#T (prop.) phi*I_a (prop.) I_sh*I_a
#math.since T_1 = T_2
I_sh2_I_a2 = I_sh1*I_a1
I_sh2_I_a2 = I_sh1*I_a1 # = 20
# Calculations
#N (prop.) E_b/I_sh
#E_b1 = V - I_a1*R_a
#Solving further for I_a2 we get I_a2**2 -500 I_a2 + 12800
p = [1, -500, 12800]
ans = roots(p)
I_a2 = ans[1] #higher root is neglected
I_sh2 = I_sh2_I_a2 / I_a2
R_x = (V / I_sh2) - R_sh #resistance to be inserted in shunt field
# Results
print 'resistance to be inserted = %.4f ohms '%(R_x)
import math
# Variables
V = 250.
N_1 = 1000.
I_L1 = 25.
R_a = 0.2
R_sh = 250. #armature and shunt field resistance
V_brush = 1. #voltage drop due to brushes
# Calculations and Results
I_sh1 = V/R_sh
I_a1 = I_L1 - I_sh1
E_b1 = V- I_a1*R_a - 2 *V_brush
#when loaded
I_L2 = 50.
I_sh2 = I_sh1 #as flux weakensby armature reaction shunt field current remains same
I_a2 = I_L2 - I_sh2
E_b2 = V- I_a2*R_a - 2 *V_brush
phi2_by_phi1 = 1- (3./100) #weakens by 3 percent
N_2 = N_1*(E_b2/E_b1)/ phi2_by_phi1 #N (prop.) E_b/phi
print 'New speed = %.3f rpm'%(N_2)
T_1 = E_b1*I_a1/(2*math.pi*N_1/60)
T_2 = E_b2*I_a2/(2*math.pi*N_2/60)
print 'Torque before field weakening = %.4f N-m'%(T_1)
print 'Torque after field weakening = %.4f N-m'%(T_2)
import math
# Variables
V = 220.
R_a = 0.5
R_x = 1. #armature resistance and extra resistance
N_FL = 500. #full load speed in r.p.m
I_a_FL = 30.
# Calculations and Results
#part(i) Full load
E_b_FL = V- I_a_FL * R_a
#T (prop.) I_a... T is constant
I_a_dash_FL = I_a_FL
E_b_dash_FL = V- I_a_dash_FL * (R_a+R_x)
#N (prop.) E_b/phi (prop.) E_b
N_dash_FL = N_FL*(E_b_dash_FL/E_b_FL)
print 'i)Speed at full load torque = %.4f r.p.m'%(N_dash_FL)
#part(ii)
T2_by_T1 = 2
I_a_dash_FL = I_a_FL *(T2_by_T1)
E_b_dash_FL = V- I_a_dash_FL * (R_a+R_x)
N_dash_FL = N_FL*(E_b_dash_FL/E_b_FL)
print 'ii)Speed at double full load torque = %.3f r.p.m'%(N_dash_FL)
#part(iii) ...stalling
E_b = 0 #as speed is zero in case of stalling torque
I_a_stall = (V-E_b)/(R_a+R_x)
T_FL = E_b_FL * I_a_FL/(2*math.pi*N_FL/60)
T_stall = T_FL *(I_a_stall/ I_a_FL)
print 'iii)Stalling torque = %.3f Nm'%(T_stall)
import math
# Variables
V = 230.
I_a1 = 30.
R_a = 0.4
R_x = 1.1 #armature resistance and extra resistance
N_1 = 500.
# Calculations and Results
#part(i)
E_b1 = V - I_a1*R_a
I_a2 = I_a1 #I_a is constant as T , phi are constant
E_b2 = V - I_a2*(R_a+R_x)
N_2 = N_1 *(E_b2/E_b1) #Because N (prop.) E_b/phi (prop.) E_b
print 'i)Speed at full load torque = %.3f r.p.m'%(N_2)
#part(ii)
T2_by_T1 = 1.5
I_a2 = I_a1 * T2_by_T1
E_b2 = V - I_a2*(R_a+R_x)
N_2 = N_1 *(E_b2/E_b1) #Because N (prop.) E_b/phi (prop.) E_b
print 'ii)Speed at 1.5 times full load torque = %.3f r.p.m'%(N_2)