Time domain equation for RC circuit with AC input

  • #1
eyeweyew
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Schoolwork-type thread moved from the techincal forums to the schoolwork forums
TL;DR Summary: How to find Time domain equation for RC circuit with AC input from inverse laplace transform

For a simple RC circuit with AC input such as this: https://www.electronics-tutorials.ws/wp-content/uploads/2013/08/rc12.gif?fit=310,151?fit=310,226. If the AC input is just a simple sinusoidal vI(t)=Vicos(ωt) and the transfer function is Vo/Vi=1/(1+sRC). The inverse Laplace transform of Vo/Vi is e-t/RC/RC. How exactly do I find the Time domain equation for vO(t) from this inverse laplace transform? I cannot seem to find any reference to show how to do that.
 
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  • #2
1705889727687.png
 
  • #3
Have you tried searching the web for this. I looked for "solve RC circuit with Laplace transforms" and got lots of links that looked good.

Is this a home work problem? At what point in the analysis are you getting confused? Can you show us what you've tried so far?
 
  • #4
If ##\frac{V_o(s)}{V_i(s)} = \frac{1}{(1+sRC)}## then ##V_o(s) = \frac{1}{(1+sRC)} V_i(s) ##

## v_o(t) = \mathcal L^{-1}(V_o(s)) = \mathcal L^{-1}(\frac{1}{(1+sRC)} V_i(s)) ##

What is ## V_i(s) ##?
 
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  • #5
I think I got it. Thanks!
 
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FAQ: Time domain equation for RC circuit with AC input

What is the time domain equation for an RC circuit with an AC input?

The time domain equation for an RC circuit with an AC input is typically derived from Kirchhoff's voltage law (KVL). For a series RC circuit with an AC voltage source \( V(t) = V_0 \sin(\omega t) \), the equation is \( V(t) = V_R(t) + V_C(t) \), where \( V_R(t) = R \cdot I(t) \) and \( V_C(t) = \frac{1}{C} \int I(t) \, dt \). Combining these, the differential equation is \( V_0 \sin(\omega t) = R \cdot I(t) + \frac{1}{C} \int I(t) \, dt \).

How do you solve the time domain equation for the current in an RC circuit with an AC input?

To solve for the current \( I(t) \) in the time domain, you need to solve the differential equation \( V_0 \sin(\omega t) = R \cdot I(t) + \frac{1}{C} \int I(t) \, dt \). This can be done using methods such as the Laplace transform, which converts the equation into the s-domain, solving for \( I(s) \), and then applying the inverse Laplace transform to find \( I(t) \). The solution typically involves both a transient and a steady-state component.

What is the steady-state solution for the current in an RC circuit with an AC input?

The steady-state solution for the current in an RC circuit with an AC input can be found by considering only the sinusoidal steady-state response. Assuming the input voltage is \( V(t) = V_0 \sin(\omega t) \), the steady-state current \( I_{ss}(t) \) can be expressed as \( I_{ss}(t) = I_0 \sin(\omega t + \phi) \), where \( I_0 \) is the amplitude and \( \phi \) is the phase shift. These can be determined using phasor analysis, resulting in \( I_0 = \frac{V_0}{\sqrt{R^2 + (\frac{1}{\omega C})^2}} \) and \( \phi = -\tan^{-1}(\frac{1}{\omega RC}) \).

How does the phase angle between voltage and current change in an RC circuit with an AC input?

The phase angle \( \phi \) between the voltage and the current in an RC circuit with an AC input is determined by the ratio of the resistive and capacitive reactances. Specifically, \( \phi = -\tan

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