PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

430 Practical MATLAB® Applications for Engineers

subplot(3,1,3)

ezplot(VR2 _ t)

title(‘VR2(t) vs. t’);axis([0 3 0 110]);

xlabel(‘time (in sec)’);ylabel(‘ Amplitude (in volts)’)

The script fi le loop_laplace_eqs is executed and the results are as follows (Figures 4.75

and 4.76):

>> loop _ laplace _ eqs

******************************************

*********** R E S U L T S *************

*********** C U R R E N T S **************

The loop currents i1(t) and i2(t) (in amps) are:

i1(t)=

• 40/3 exp(- 5/3 t) + 20
  i2(t)=
  20/3 exp(- 5/3 t)

  ---

  
  

  The currents i1(t=0) and i2(t=0) (in amps)
  using the initial value theorem are:
  i1 0 =
  20/3
  i2 0 =
  20/3

  
  

  ---

  
  

  The current i1(t=inf) and i2(t=inf) (in amps)
  using the final value theorem are :
  i1 inf =
  20
  i2 inf =
  0

  
  

************ V O L T A G E S****************

The voltage across the inductor L (in volts) is given by:

VL _ t =

6*exp(-5/3*t)+10*t-6

The voltage across the resistor R1 (in volts) is given by:

VR1 _ t =

-200/3*exp(-5/3*t)+100

The voltage across the resistor R2 (in volts) is given by:

VR2 _ t =

200/3*exp(-5/3*t)

Note that at t = 0, the inductor acts as an open circuit, then i 1 ( 0 ) = i 2 ( 0 ) = 100 V/ 15 Ω =

20/3 A; and at t = ∞, the inductor becomes a short circuit, then i 1 (∞) = 100 V/ 5 Ω = 20 A,

and i 2 (∞) = 0 A. Values that completely agree with the results obtained using the

MATLAB symbolic toolbox.