Circuit Theory/Complex Frequency Examples/example12

A voltage source generates a current when connected to an RC branch. Find the values of the resistance and capacitance.

${\displaystyle v(t)=10e^{-t}sin(7t)}$
${\displaystyle i(t)=0.2e^{-t}cos(7t-30^{\circ })}$

knowns and unknowns

knowns ${\displaystyle v(t)}$  and ${\displaystyle i(t)}$
unknowns ${\displaystyle RC}$

equations

${\displaystyle v_{R}=i*R}$
${\displaystyle i=C*{d \over dt}v_{C}}$
${\displaystyle V_{s}=v_{C}+v_{R}}$

into complex frequency domain symbolically

${\displaystyle \mathbb {V} _{R}=\mathbb {I} *R}$
${\displaystyle \mathbb {I} =C*s*\mathbb {V} _{C}}$
${\displaystyle \mathbb {V} _{s}=\mathbb {V} _{R}+\mathbb {V} _{C}}$

Solving symbolically as much as possible:

${\displaystyle \mathbb {V} _{s}=\mathbb {I} *R+{\frac {\mathbb {I} }{C*s}}}$
${\displaystyle {\frac {\mathbb {V} _{s}}{\mathbb {I} }}=R+{\frac {1}{C*s}}}$
${\displaystyle s=\phi +j\omega }$

At this point, symbolic manipulation with mupad fails. Mupad can not separate out symbolically the rectangular or polar expressions. Doing this by hand will result in huge symbolic expressions.

It is easier just to leave the solution in the above symbolic form.

At this point, it seems like there is one equation with two unknowns (R, C). But in reality there are two equations. The real part of the right has to equal the real part on the left. The imaginary on the right has to equal the imaginary on the left. Unfortunately, the values of ${\displaystyle \mathbb {I} }$  and ${\displaystyle \mathbb {V} _{s}}$  will determine the next step.

find RC of series circuit given Vs and Is .. m file

complex domain math

${\displaystyle s=-1+j7}$
${\displaystyle \mathbb {V} _{s}=-j10}$
${\displaystyle \mathbb {I} =0.2*cos(-30^{\circ })+j*0.2*sin(-30^{\circ })=0.1*{\sqrt {3}}-j0.1}$
${\displaystyle {\frac {\mathbb {V} _{s}}{\mathbb {I} }}=25-43.3j}$
${\displaystyle R+{\frac {1}{C*s}}=R+{\frac {-0.14*i-0.02}{C}}}$
Thus ${\displaystyle -43.3j={\frac {-0/14j}{C}}}$
${\displaystyle C={\frac {-0.14j}{-43.3j}}=3.233mF}$
${\displaystyle 25=R+{\frac {-0.02}{C}}}$
${\displaystyle R=25+{\frac {0.02}{.003233}}=31.19ohms}$

Solution can only be found in the complex domain.

Simulation

Can not simulate to get the answer, but can simulate the answer to see if matches the sources given.

${\displaystyle period={\frac {2*\pi }{\omega }}=0.8976sec}$

It is clear that the cos, current is leading the voltage. Theoretically should be by -30 + 90 = 60 degrees.

The difference between the peaks is 0.141 seconds approximately. This translates in to degrees:

phase difference = peak difference * 360/period = 56.55 degrees .. approximately 60 degrees.

Magnitude of the current is less than the 0.2 specified, but is close ... approximately 0.18 amps. There is lots of error here. Probably could do more accurately.