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For this circuit I need to find \$V_C(0)\$, \$V_C(\infty)\$, \$i_L(0)\$, \$i_L(\infty)\$, \$V_C(t)\$ when \$t>0\$ and \$I_C(t)\$ when \$t>0\$.

schematic

simulate this circuit – Schematic created using CircuitLab

So far, I have

  • \$V_C(0) = 30\ \mathrm{V}\$
  • \$V_C(\infty) = 30\ \mathrm{V}\$
  • \$i_L(0) = 10\ \mathrm{A}\$
  • \$i_L(\infty) = 0\ \mathrm{A}\$

To find \$V_C(t)\$, I used $$ Vc(t) = V_o\cdot e^{-\tfrac{t}{RC}} = 30 e^{-\tfrac{t}{1.5}}. $$ (I don't think this is the right equation, but I don't know what other equation to use).

To find \$i_L(t)\$, I first found that \$a= R/2L = 3/2\$ and \$W_o = 1/\sqrt{LC} = 1/\sqrt{1/2} = \sqrt{2}\$.

Since \$a > Wo\$, the circuit is overdamped, so I'm using the $$ i_L(t)= A_1e^{s_1\cdot t} + A_2e^{s_2\cdot t}. $$

To find \$s\$ values I used \$-a \pm \sqrt{a^2-W_o^2}\$, obtaining \$s_1 = -1\$ and \$s_2 = -2\$.

Plugging that in, I now have \$i_L(t) = A_1e^{-t} + A_2e^{-2t}\$.

To start solving for \$A_1\$ and \$A_2\$ I set \$t=0\$ so I have \$10 = A_1 + A_2\$.

I don't know how to get another equation relating \$A_1\$ and \$A_2\$ so I'm stuck here.

I thought I was understanding \$RLC\$ circuits, but I got stuck and now I'm not sure that I have any of this right. Could someone please look through my work and see where I went wrong?

ocrdu
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offtoco
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1 Answers1

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You actually don't need any equations for this, it's testing your intuitive understanding of RLC circuits.

  • The voltage across an inductor after a long time is zero.
  • The current through a capacitor after a long time is zero.
  • The voltage across a capacitor does not change instantly.
  • The current through an inductor does not change instantly.

If you use applicable rules as above and Ohm's law your answer should be clear.

These apply provided there is some resistance in the circuit somewhere, otherwise the current could slosh back and forth forever, in the simplified way of looking at things.

Once you have the initial conditions you can apply the equations for an RLC circuit.

Spehro Pefhany
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