Question

If switch is closed at t=0,what would be Thevenin equivalent transform voltage at terminals a-a' ?(see attachment)

Answer 1

its a series LC circuit ,

damping coeffiecent = 0

resonant frequency = 1/ sqrt(LC)

Answer 2

resonant frequency = 1/ sqrt( 1/ 25) = 1 / (1/5) = 5rad/sec

Answer 3

solves of the characteristic equations are \[s=-\alpha+- \sqrt{\alpha^2-\omega^2}\]

Answer 4

so solutions of the characteristic eqn are +-5i , complex conjugates

so the general solution for the inductor current ( because then we can find the voltage by differentiating )

is i(t) = e^(0t) ( Acos(5t) +Bsin(5t) ) = Acos(5t) +Bsin(5t)

Answer 5

Note that the final value of the current is zero

Answer 6

now we just need the initial conditions to find the constants

i(0) = 0 ( because the circuit is disconnected )

Answer 7

then when the switch is closed, do a KVL around the loop

10 - Vc(0) -VL(0) =0

Answer 8

but VL = L di/dt = 4 di/dt

also Vc(0) is the initial capacitor voltage

Answer 9

the initial capacitor voltage is unknown.

when the circuit is at steady state the inductor is a short circuit , so the RHS of the capacitor is at ground, 0V , but the LHS is at an open switch , and an open circuit doesn't always mean 0V.

Answer 10

O.K.
I found that V(s)=10s/s^2+25.
Am I right ?

Answer 11

capacitor becomes 1/[ s (1/100) ] = 100/s

inductor becomes 4s

Answer 12

voltage divider

Answer 13

\[\frac{4s}{4s + \frac{100}{s}} \times 10\]=

Answer 14

whatever that is

Answer 15

That is V(S)=10s^2/s^2+25,You are right.
Thank you for your time !!!!!!