In the circuit below (working in the Laplace domain), I have to calculate \$I\$.

$$\begin{align} &I=\frac{-V_L-V_C}{sL+\frac{1}{SC}}=\frac{-V_L - \frac{V_0}{s} - I \frac{1}{sC} }{sL+\frac{1}{SC}}\implies\\ &\implies\ I\Big(1+\frac{\frac{1}{sC}}{sL+\frac{1}{sC}}\Big)= \frac{-V_L-\frac{V_0}{s}}{sL+\frac{1}{sC}}\implies\\ &\implies I=\frac{sL+\frac{1}{sC}}{sL+\frac{2}{sC}} \frac{-V_L-\frac{V_0}{s}}{sL+\frac{1}{sC}}=\frac{-V_L-\frac{V_0}{s}}{sL+\frac{2}{sC}} \end{align}$$

By seeing the solution on my book, it gives $$I=\frac{\frac{V_0}{s}}{sL+\frac{1}{sC}}$$ Why is that? Where am I wrong?


simulate this circuit – Schematic created using CircuitLab


2 Answers 2


Your first equation is wrong, or I don't grasp your notation. I equals the applied voltage divided by the impedance, so the first formula you use is not quite right. The third term if different from the second, and should make zero since you are summing all the voltages along a closed loop.

The proper way to tackle this problem is apply ohm's law:

$$ V = I\cdot R \rightarrow I = \frac{V}{R} $$

The law is still valid in the s domain if you use the impedances:

$$ I(s) = \frac{V(s)}{Z(s)} $$

In your circuit you have an impedance that is the series of a capacitor and an inductor, thus: $$ Z(s) = sL + \frac{1}{Cs} $$

While \$V(s)\$ is given. Finally: $$ I(s) = \frac{\frac{V_0}{s}}{sL + \frac{1}{Cs}} $$

You can have a little more fun and come to: $$ I(s) = V_0\frac{C}{s^2LC+1} $$

  • \$\begingroup\$ Shouldn't $V(s)$ be the voltage between $A$ and $B$? \$\endgroup\$
    – sl34x
    Commented May 20, 2015 at 12:14
  • 1
    \$\begingroup\$ The voltage between A and B is 0 since they are the same node. Before working in the s domain you should maybe review your 'circuit solving' class (or however you call it). \$\endgroup\$ Commented May 20, 2015 at 12:16
  • \$\begingroup\$ Yes, you're right. I didn't figure it out. I redrew the circuit on a piece of paper and I figured it out in the right way. Thanks a lot, anyway! \$\endgroup\$
    – sl34x
    Commented May 20, 2015 at 12:40

Basically,if the IC'S for inductance is ZERO, we transfer all elements of the given circuit from t-domain to S-domain, therefore, from the S-domain circuit we get the following equation by applying KVL - current I(s) counterclockwise, hence, +(Vo/s) - ((LS) I(S)) - (1/CS) I(S) = 0 Consequently, I(s) required is equal (VO/S)/((LS)+(1/CS)). Note, if the inductance has IC'S then you should add a voltage source in series with the inductance and do the same method as done early adding the Li(0) to the equation written above, which gives I(s) = ((VO/S)+(LI(0))/((LS)+(1/CS)).


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