Book
Advanced Engineering Mathematics
Author
Erwin Kreyszig
Edition
8th Edition
#We wil use L for Inductance, C for capacitance, R for Ohm's resistance.
L,R,t,E0 = var('L,R,t,E0') # Defining the variables we will be using
E = function('E',t) # E is function of t
I = function('I',t) # I is function of t
show(L*diff(I,t) + R*I -E)
ode_model = L*diff(I,t)+R*I-E # This is a ordinary differential equations we have to solve
#First we will do it for simple case where L = 0.1 henry, R = 5 ohms and V = 12 volts
E(t) = 12
ode_model1 = L*diff(I,t)+R*I-E
ode_model2 = ode_model1.subs(L=0.1,R=5) # Substituting the values of L and R
ode_sol = desolve(ode_model2,I) # This is a solution of ode
show(ode_sol)
# Variation of graph with respect to variation in initial condition
@interact
def f(i1=(-2.4,2.4,0.4)):
ode_sol1 = ode_sol(t,c=i1)
show(plot(ode_sol1,t,0,0.1,axes=true, xmin=0, xmax=0.15, ymin=0, ymax=5))
E(t) = E0
ode_model1 = L*diff(I,t)+R*I-E
ode_sol = desolve(ode_model1,I,ivar=t) # This is a solution of ode
show(ode_sol)
#Now we will do it for simple case where L = 0.1 henry, R = 5 ohms and E = E0sin(wt)
E0,w = var('E0,w')
E(t) = E0*sin(w*t)
ode_model1 = L*diff(I,t)+R*I-E
#ode_model2 = ode_model1.subs(L=0.1,R=5) # Substituting the values of L and R
ode_sol = desolve(ode_model1,I,ivar = t) # This is a solution of ode
print(ode_sol.simplify())
# Will plot the graph for a simplified case, when the assumptions are made.
I(t)=exp(-0.1*t) + sin(t-pi/4)
plot(I(t),t,0,12*pi)
Solution by:
- Praveen Kumar,Student,IIT Roorkee
- Tale Praffullkumar, Student, IIT Roorkee