anyone knows use Laplace to find ZT and IT?

Ratch

Joined Mar 20, 2007
1,070
kuangmao,

and please show me the work steps
Your result Zt = (2*(3*s^2+14*s+7))/(s^2+5*s+2) is correct

Dividing Zt into voltage 6/s gives I(s) = (3*(s^2+5*s+2))/(s*(3*s^2+14*s+7))

Converting I(s) to partial fractions gives I(s)=(1/7)*(3*s+21)/(3*s^2+14*s+7)+6/(7*s)

From a table in Laplace transforms we find that s/(s^2-a^2)=(sinh(t))/a and s/(s^2-a^2)= cosh(t), so the inverse Laplace gives 6/7+(1/7)*exp(-(7/3)*t)*(cosh((2/3)*t*sqrt(7))+sqrt(7)*sinh((2/3)*t*sqrt(7)))

Ratch
 

Ratch

Joined Mar 20, 2007
1,070
kuangmao,

Enclosed in the attachment is the solution in the time domain in hyperbolic format already given and exponential format. Both are equivalent as proven by the identical plots also shown. You will notice that after a long time the current settles into a steady state value of 6/7 amps, which can be ascertained by an inspection of the schematic. This is also shown on the plots.

Ratch
 

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