Tokoro

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Thanks Kitonum 19099 and sursumCorda 20.

Both methods solve this puzzle symbolically. 

Tokoro.


 

:RLC circuit I want to solve.

restart

with(DynamicSystems); with(inttrans)

E := 1; J := 3; R := 3; L := 1; C := 1/2

(1/s-6/s^2)/(s+3+2/s)

(1/s-6/s^2)/(s+3+2/s)

(1)

i__RL := proc (t) options operator, arrow; invlaplace((1/s-6/s^2)/(s+3+2/s), s, t) end proc

proc (t) options operator, arrow; invlaplace((1/s-6/s^2)/(s+3+2/s), s, t) end proc

(2)

"(->)"

-4*exp(-2*t)-3+7*exp(-t)

(3)

e__C := proc (t) options operator, arrow; E-i__RL(t)*R-L*(diff(i__RL(t), t)) end proc

proc (t) options operator, arrow; E-i__RL(t)*R-L*(diff(i__RL(t), t)) end proc

(4)

"(->)"

10+4*exp(-2*t)-14*exp(-t)

(5)

plot([-4*exp(-2*t)-3+7*exp(-t), 10+4*exp(-2*t)-14*exp(-t)], t = 0 .. 10, color = [red, blue], thickness = 3, labels = ["t", "i(t)_red and vc(t)_blue"])

 

NULL

(1/s-6/s^2)/(s+3+2/s)

(1/s-6/s^2)/(s+3+2/s)

(6)

Solve eq(6) by Maple's dsolve.

E := 1; R := 3; C := 1/2; L := 1; J := 3

sol := dsolve([L*(diff(diff(q(t), t), t))+R*(diff(q(t), t))+q(t)/C = 1-6*t, q(0) = 0, (D(q))(0) = 0])

q(t) = 2*exp(-2*t)-7*exp(-t)-3*t+5

(7)

rhs(q(t) = 2*exp(-2*t)-7*exp(-t)-3*t+5)

2*exp(-2*t)-7*exp(-t)-3*t+5

(8)

plot(diff(2*exp(-2*t)-7*exp(-t)-3*t+5, t), t = 0 .. 10, color = red, thickness = 3)

 

v[c] := proc (t) options operator, arrow; E-R*(diff(2*exp(-2*t)-7*exp(-t)-3*t+5, t))-L*(diff(2*exp(-2*t)-7*exp(-t)-3*t+5, t, t)) end proc

proc (t) options operator, arrow; E-R*(diff(2*exp(-2*t)-7*exp(-t)-3*t+5, t))-L*(diff(2*exp(-2*t)-7*exp(-t)-3*t+5, t, t)) end proc

(9)

(proc (t) options operator, arrow; E-R*(diff(2*exp(-2*t)-7*exp(-t)-3*t+5, t))-L*(diff(2*exp(-2*t)-7*exp(-t)-3*t+5, t, t)) end proc)(t)

10+4*exp(-2*t)-14*exp(-t)

(10)

i__c := proc (t) options operator, arrow; diff(C*(10+4*exp(-2*t)-14*exp(-t)), t) end proc

proc (t) options operator, arrow; diff(C*(10+4*exp(-2*t)-14*exp(-t)), t) end proc

(11)

(proc (t) options operator, arrow; diff(C*(10+4*exp(-2*t)-14*exp(-t)), t) end proc)(t)

-4*exp(-2*t)+7*exp(-t)

(12)

plot([10+4*exp(-2*t)-14*exp(-t), -4*exp(-2*t)+7*exp(-t)], t = 0 .. 5, color = ["red", "blue"], thickness = 3, labels = ["t", "vc(t)_red,ic(t)_blue"])

 

NULL

NULL


 

Download DynamicSystem-How_to_ODE.mw

@Preben Alsholm It makes equation (7). And I want to solve this ODE in Eq.(7) by Maple.

Using laplace convertion, I can get the answer and can plot it.

@acer Thanks show me the power of Maple. 

@acer Thanks your follow up. I try to find the mimimum of the intersect. 

Optimization can find it. How to find by min(extrema....? 

 

Totoro.

plot-15-iPlot.mw 

@tomleslie x=45, y=45, intersect=2.117 may be the maximum of intersect. How to find this value by Maple?

The only one 3D mesh answer for "Make one ohm by one ohm."

The only one 2D mesh for "Make one ohm by one ohm".

Node 1 to 13 is 1/2 ohm made by one ohm, where each current is different.

 

@dharr @Carl Love is right. Surprisingly, both are planar. I only know this very fameous math problem. I only know the answers and don't know how to solve. I rewrited this math problem to the electric circuit problem.

Tokoro

Same question except make two ohm by one ohms, each current must different.

@dharr Yes, "all currents are different and none of them are zero".
Please try to calculate the resistance of node 1 to 13. This is very good electrical engineering problem that without using computer math software, such as Maple, not only students but also professors can't solve.

Maple new user. Tokoro.