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The circuit is closed At t = 0 , initial state of capacitor v(0-) = 1v and inductor i(0-) = 0A.

My problem is when I convert this circuit into Laplace domain resistor become 2 and inductor become S. What happen to capacitor.

Please give an advice. Thank you.

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  • \$\begingroup\$ Where's your differential equation? \$\endgroup\$
    – Andy aka
    Commented Sep 9, 2021 at 7:07
  • \$\begingroup\$ @Andyaka I didn't write differential equation because if I can directly this circuit convert to Laplace domain then capacitor, inductor become resistors so then easily I can use KVL to Laplace domain circuit but I don't understand to convert capacitor with initial state. \$\endgroup\$
    – Chathura1
    Commented Sep 9, 2021 at 7:11
  • \$\begingroup\$ Write Laplace equation first, without replacing values. this can help you. dummies.com/education/science/science-electronics/… \$\endgroup\$
    – Antonio51
    Commented Sep 9, 2021 at 7:59
  • \$\begingroup\$ The impedance of a capacitance C is 1/(sC) in the frequency domain. \$\endgroup\$
    – Bart
    Commented Sep 9, 2021 at 8:32
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    \$\begingroup\$ @Chathura1 Are you supposed to solve this as a time domain result? Looks like it to me. But I'm not entirely sure. \$\endgroup\$
    – jonk
    Commented Sep 9, 2021 at 9:57

1 Answer 1

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To take into account the initial voltage of a charged capacitor, only add this voltage (Vo/s, of course) as a additional serial source (with capacitor) in your circuitry.

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Here is an illustration what happens in 3 similar problems. At time only framed in green.

enter image description here

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