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We know that current through capacitor is i(t)=c*dv(t)/dt but what if we want the current through capacitor expressed in Laplace form ?

schematic

simulate this circuit – Schematic created using CircuitLab

Here we convert 2F into Laplace as 1/2*s but what after then? Assume that i current flows through capacitor .

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2 Answers 2

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At least two ways of looking at this:

  1. The Laplace representation of the capacitor's reactance is \$\frac{1}{sC}\$, hence for a voltage, \$\small V(s)\$ across \$\small C\$, the current through \$\small C\$, by Ohm's law, will be \$\small I(s)=sC\:V(s)\$

  2. Differentiation in the time domain is equivalent to multiplying by \$\small s\$ in the Laplace domain. Therefore \$\frac{dv}{dt}\small \rightarrow sV(s)\$, and the differential equation transforms to: \$\small I(s) = C\:sV(s)\$

Zero initial conditions have been assumed.

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  • \$\begingroup\$ yes this was what i was looking for \$\endgroup\$ Commented Jun 12, 2016 at 22:46
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Remember that s = \$2\pi f\$ so if the frequency is 1 Hz then the impedance is \$\dfrac{1}{2\times 2\pi \times 1}\$ = 0.0796 ohms (reactive).

If the applied voltage is a 1 Hz sinewave of amplitude 1 V RMS then current is 12.57 amps.

If the applied current is a 1 Hz sinewave of amplitude 1 A RMS then the voltage will be 0.0796 volts.

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