1
\$\begingroup\$

I would like some help with Sadiku's exercise about second order circuits.

I did'nt understand why the solution says that V (0) = -30 V.

Also, I did not understand why the capacitor's voltage (Vc) is -30 V.

Isn't the voltage for t <0 (when the capacitor is loaded) the same as the voltage source?enter image description here

The exercise asked i(t) for t>0 . The switch opens at t=0 after a long time closed.

enter image description here


If I change to + - in the capacitor as suggested.

For t<0: V(0)= 30 V and i(0)= 15 A

For t>0:

α= R/(2L)= 2/(2*0,5)= 2

omega = 1/sqrt (LC) = 1/sqrt(0.5*0.25) = 2.82942712475

Under-damped:

di/dt= -2(15 cos(2t)+ B sin(2t))e^-2t + (-2*15 sin (2t) + 2 Bcos(2t))e^2t

di(0)/dt = -30+ 2B = -(1/L) [R*i(o)+ Vc(0)] = -2 [30+30] = -120

-30+2B=-120

B= -45

i(t)= 15 cos (2t) + (-45 sin (2t))e^-2t A

So, the answer isn't the same as the book's solution.

\$\endgroup\$
  • \$\begingroup\$ When googling Sadiku, only football guys are coming up. Are we supposed to know who it is? \$\endgroup\$ – Eugene Sh. Dec 17 '18 at 20:37
  • \$\begingroup\$ Sorry, it's a famous book on Eletrical Engeneering graduation. It's called Fundamentals of Electric Circuits, 5th edition by Matthew Sadiku \$\endgroup\$ – Bernardri Dec 17 '18 at 20:41
  • 1
    \$\begingroup\$ Why do you think the answer is wrong? What is your version? The capacitor is fully charged initially, so it's voltage equals to the one of the power supply. Since the polarity of it is defined opposite to the one of the supply, it has the negative sign. \$\endgroup\$ – Eugene Sh. Dec 17 '18 at 20:43
  • \$\begingroup\$ According to my researches the voltage at the capacitor is the same as the voltage source if the capacitor is loaded. So, V(0) for t<0 should be equal to 30 V. (Voltage Source). Also the capacitor should have 30 V, and the solution book says that the voltage Vc(0) = -30 V. \$\endgroup\$ – Bernardri Dec 17 '18 at 20:48
  • \$\begingroup\$ Have you read my whole comment? Vc is defined in opposite polarity than the supply, as shown on the second drawing. \$\endgroup\$ – Eugene Sh. Dec 17 '18 at 20:49

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.