Metodos Numericos

V_t-V_R-V_L=0
V_t=V_R+V_L
V_t=IR+L dI/dt
dI/dt=V_t/L-R/L I(t)
Formula
y_(i+1)=y_i+(x_i,y_i )h
I_(i+1)=I_i+(t_i,I_i )h
Desarrollo
dI/dt=12/25-120/25 I
dI/dt=0.48-4.8I
I_(i+1)=I_i+(t_i,I_i)h
I_1=I_i+(0.48-4.8I)h
I_1=0+(0.48-4.8(0))0.1
I_1=0+(0.48)0.1
I_1=0.048
I_2=I_i+(0.48-4.8I_i )h
I_2=0.048+(0.48-4.8(0.048))0.1
I_2=0.048+(0.48-0.2304)0.1
I_2=0.048+(0.2496)0.1I_2=0.048+0.02496
I_2=0.07296
I_3=I_i+(0.48-4.8I_i )h
I_3=0.07296+(0.48-4.8(0.07296))0.1
I_3=0.07296+(0.48-0.38208)0.1
I_3=0.07296+(0.129792)0.1
I_3=0.07296+9.792〖x10〗^(-3)
I_3=0.0859392
I_4=I_i+(0.48-4.8I_i)h
I_4=0.0859392+(0.48-4.8(0.0859392))0.1
I_4=0.0859392+(0.48—0.41250816)0.1
I_4=0.0859392+(0.06749184)0.1
I_4=0.0859392+6.7491804〖x10〗^(-3)
I_4=0.092688384
I_5=I_i+(0.48-4.8I_i )hI_5=0.092688384+(0.48-4.8(0.092688384))0.1
I_5=0.092688384+(0.48—0.4449042432)0.1
I_5=0.092688384+(0.0350957568)0.1
I_5=0.092688384+3.50957568〖x10〗^(-3)
I_5=0.09619795968
i ti Ii Erp
0 0 0 100
1 1 0.4834.2105263
2 2 0.7296 15.1027703
3 3 0.859392 7.28158558
4 4 0.92688384 3.64828494
5 5 0.9619796

I_i analitica=(E5:E9)+(0.48-(4.8*(E5:E9)*0.1))
Erp=ABS(((E6:E10)-(E5:E9))/(E6:E10))*100dI/dt=V_t/L-R/L I(t)
dI=(V_t/L-R/L I)dt
∫▒dI=∫▒(V_t/L-R/L I)dt
∫▒dI=V_t/L ∫▒dt-R/L I∫▒dt
I=V_t/L t-RI/L t+c……………………………….Ec 1
Sustituyendo en ecuación 1
V_t=12 v L=25 HR=120Ω
I(0)=0 t=0
0=12/25(0)-(120)(0)/2(25) (0)+c
c=0
Sustituir c en Ecuacion 1
I=V_t/L t-RI/2L t+0
I=12/25 t-12I/5 t
I=0.48t-2.4It
2.4It+I=0.48t
I(2.4t+1)=0.48tI=0.48t/((2.4t+1) )……………………..Ec2
Sustituir t=1en Ec 2
I_1=0.48t/((2.4t+1) )
I_1=0.48(1)/((2.4(1)+1) )
I_1=0.48/((2.4+1) )
I_1=0.48/3.4
I_1=12/85
Sustituir t=2en Ec 2
I_2=0.48t/((2.4t+1) )I_2=0.48(2)/((2.4(2)+1) )
I_2=0.096/((4.8+1) )
I_2=0.096/5.8
I_2=12/725
Sustituir t=3en Ec 2
3=0.48t/((2.4t+1) )
I_3=0.48(3)/((2.4(3)+1) )
I_3=1.44/((7.2+1) )
I_3=1.44/8.2
I_3=36/205
Sustituir t=4en Ec 2...