PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

Direct Current and Transient Analysis 191

5. By using MATLAB, verify the solution vC(t) obtained in part 2 by applying KVL
   around the active loop for t > 0.


6. Compare the results of part 1 with the results of part 2 and 3.

ANALYTICAL Solution

The loop differential equation for t ≥ 0 is given by

20 
vtC() 2 it() for 0t

with νC(0) = 10 V and νC(∞) = 20 V.

Replacing i(t) = C

dvc(t)

_____dt in the preceding differential equation yields

220

dv t

dt

C vtC

()


()

Then assuming a solution for vC(t) of the form vC(t) = A + B e−t/τ, where τ = C (^) R = 2 s,
and knowing the initial and fi nal voltage values across C, the constants A and B can be
evaluated. By following this process the voltage across C is given by vC(t) = 20 − 10 e−0.5t V,
the voltage across R is vR(t) = 10 e−0.5t V, and the current is i(t) = 5 e−0.5t A.
MATLAB Solution
% Script file: RC _ IC
syms t;
% symbolic solution
vct=dsolve(‘2Dy+y=20’,’y(0)=10’,’t’);
disp(‘*****’);
disp(‘*R E S U L T S *’);
disp(‘***’);
FIGURE 2.92
Network of Example 2.27.
V 1 = 10 V
Switch moves down at t = 0

• −
  C 1 =1 F
  R = 2 Ω
  V 2 = 20 V
  i(t)