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I'm trying to learn kirchhoff's law. but im stuck on how to solve for a circuit that has both a resistor and a capacitor. How would you proceed?

Example Question

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I'd start by writing down Kirchoff's voltage law. Then, I'd think about how to calculate the voltage for each element in the circuit and insert those into Kirchoff's voltage law.

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  • $\begingroup$ I tried that, but i get stuck once I get this equation 9v = I(0.5)+Ve^(-t/RC). How would I continue? $\endgroup$ – Stan Feb 23 '19 at 21:32
  • $\begingroup$ OP, you're not using the right expression for voltage across a capacitor. You want one that's fundamental to the nature of the capacitor, not one that depends on the circuit it's in. $\endgroup$ – The Photon Feb 23 '19 at 23:34
  • $\begingroup$ @ThePhoton is right, the expression you're using is for an entire circuit. Start with the basic expression for the voltage across the capacitor. $\endgroup$ – HS-nebula Feb 24 '19 at 0:53
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it seems that you are passing a DC voltage to a capacitor and capacitors offer infinite impedance or resistance to DC current. The Impedance is given by Z = 1 / (omega x C) where omega is the frequency of the source if DC it is zero and hence the circuit is shorted. A fully charged Capacitor has infinite Impedance since w = 0 in the case above.

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You will have to use a bit of calculus here.

Applying Kirchhoff's Law, we get V=IR+Q/C. Now, write I=dQ/dt. After rearranging you get,

dQ/(CV-Q)=dt/(RC). Integrating with proper limits, you get ln(CV-Q)=-t/RC.

From the above relation, you got the charge on the capacitor as a function of time.By differentiating w.r.t time, you will get current through the circuit which I leave up to you.

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  • $\begingroup$ If you want to be Mathematically corrent notice that: $$\int\frac{1}{x}\space\text{d}x=\ln\left|x\right|+\text{c}$$ $\endgroup$ – Jan Feb 24 '19 at 15:28
  • $\begingroup$ You are free to edit it. $\endgroup$ – Sanket J H Mar 3 '19 at 14:53
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Well, when we have a series RC-circuit we can use Laplace transform to analyse it in detail. Using Faraday's law we can write:

$$\text{v}_\text{s}\left(t\right)=\text{v}_\text{R}\left(t\right)+\text{v}_\text{C}\left(t\right)\tag1$$

Using the relations of the voltage and current in a resitor and a capacitor we can rewrite equation $(1)$ as follows:

$$\text{v}_\text{s}'\left(t\right)=\text{i}_\text{R}'\left(t\right)\cdot\text{R}+\text{i}_\text{C}\left(t\right)\cdot\frac{1}{\text{C}}\tag2$$

Because it is a series circuit we know that the input current, $\text{i}_\text{in}\left(t\right)$, is the same as the current trough the resistor and the capacitor so we can write:

$$\text{v}_\text{s}'\left(t\right)=\text{i}_\text{in}'\left(t\right)\cdot\text{R}+\text{i}_\text{in}\left(t\right)\cdot\frac{1}{\text{C}}\tag3$$

Using the Laplace transform and assuming that the intial conditons are equal to $0$ we can write for equation $(3)$:

$$\text{s}\cdot\text{V}_\text{s}\left(\text{s}\right)=\text{s}\cdot\text{I}_\text{in}\left(\text{s}\right)\cdot\text{R}+\text{I}_\text{in}\left(\text{s}\right)\cdot\frac{1}{\text{C}}\space\Longleftrightarrow\space\text{I}_\text{in}\left(\text{s}\right)=\frac{\text{s}\cdot\text{V}_\text{s}\left(\text{s}\right)}{\text{s}\cdot\text{R}+\frac{1}{\text{C}}}\tag4$$

Writing the supply voltage in the s-domain we get:

$$\text{V}_\text{s}\left(\text{s}\right)=\int_0^\infty\hat{\text{u}}\cdot e^{-\text{s}t}\space\text{d}t=\frac{\hat{\text{u}}}{\text{s}}\tag5$$

So, for the input current we get:

$$\text{I}_\text{in}\left(\text{s}\right)=\frac{\text{s}}{\text{s}\cdot\text{R}+\frac{1}{\text{C}}}\cdot\frac{\hat{\text{u}}}{\text{s}}=\frac{\hat{\text{u}}}{\text{s}\cdot\text{R}+\frac{1}{\text{C}}}\tag6$$

So, the voltage across the capacitor is given by:

$$\text{V}_\text{c}\left(\text{s}\right)=\frac{1}{\text{s}\cdot\text{C}}\cdot\frac{\hat{\text{u}}}{\text{s}\cdot\text{R}+\frac{1}{\text{C}}}\tag7$$

And for the voltage across the resistor we get:

$$\text{V}_\text{R}\left(\text{s}\right)=\text{R}\cdot\frac{\hat{\text{u}}}{\text{s}\cdot\text{R}+\frac{1}{\text{C}}}\tag8$$

Using your values: $\text{R}=500\cdot10^{-3}\space\Omega$, $\text{C}=2\cdot10^{-6}\space\text{F}$ and $\hat{\text{u}}=9\space\text{V}$ we get (using inverse Laplace transform):

  • $$\text{i}_\text{in}\left(t\right)=18\exp\left(-1000000t\right)\tag9$$
  • $$\text{v}_\text{C}\left(t\right)=9\cdot\left(1-\exp\left(-1000000t\right)\right)\tag{10}$$
  • $$\text{v}_\text{R}\left(t\right)=9\exp\left(-1000000t\right)\tag{11}$$

Plotting the solution, using Mathematica gives:

enter image description here

Where the blue curve is the input voltage, the orange curve is the voltage across the resistor, the green curve is the voltage across the capacitor and the red curve is the input current (also the current trough the components).

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