Question

Question 1 25 pts Given the following information about a 2nd order transient circuit, solve for the coefficienta in the capacitor voltage vc (t) = e-Sunt (aj sin(wat) + a2 cos(wat)) + vc (00) for t>O. Wa=wn 1-8 C-3 mF Vcl0+) = 9V Vclinfinity) = 3V ic(O+) = 2 A wn - 82 rad/sec zeta -0.73 Question 2 25 pts

Answer 1

Answer -2

   Given, voltage across the capacitor,

     
velt) = e-Wnt (ansin(wat) + ancos(wat)) + vc
......................(1)

Given,

   C = 3 mF

   V_c(0⁺) = 9 V

   V_c($\small \infty$) = 3 V

i_c(0⁺) = 2 A

   W_n = 82 rad/sec

     
<= 0.73

Putting t = 0 in equation (1),

  $\small v_{c}(0^{+})=\alpha _{2}+v_{c}(\infty )$

  
0) - (+0Pa = 10

     $\small \alpha _{2}=9-3$

   $\small \alpha _{2}=6$

Now, the current through the capacitor is,

     $\small i_{c}(t)=C\frac{\mathrm{d} v_{c}(t)}{\mathrm{d} t}$

$\small i_{c}(t)=C\frac{\mathrm{d} }{\mathrm{d} t}\left ( e^{-\xi \omega _{n}t}\left ( \alpha _{1}sin(\omega _{d}t)+\alpha _{2}cos(\omega _{d}t) \right )+v_{c}(\infty ) \right )$

     $\small i_{c}(t)=C\left ( e^{-\xi \omega _{n}t}\left ( \alpha _{1}\omega _{d}cos(\omega _{d}t)-\alpha _{2}\omega _{d}sin(\omega _{d}t) \right )-\xi \omega _{n} e^{-\xi \omega _{n}t}\left ( \alpha _{1}sin(\omega _{d}t)+\alpha _{2}cos(\omega _{d}t) \right ) \right )$

Putting t = 0,

$\small i_{c}(0^{+})=C\left ( \alpha _{1}\omega _{d}-\xi \omega _{n} \alpha _{2} \right ) \right )$

Now,

   $\small \omega _{d}=\omega _{n}\sqrt{1-\xi ^{2}}$

     $\small \omega _{d}=(82)\times \sqrt{1-(0.73) ^{2}}$

$\small \omega _{d}=56.04(rad/sec)$

Now,

   $\small i_{c}(0^{+})=C\left ( \alpha _{1}\omega _{d}-\xi \omega _{n} \alpha _{2} \right ) \right )$

   $\small 2=(3\times 10^{-3})\left ( 56.04\alpha _{1}-(0.73)(82) (6) \right ) \right )$

     $\small \alpha _{1}=18.31$