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For a LCR circuit connected to AC source of emf$$E= e\sin(ωt)$$ and let the current in LCR circuit be I then $$I=i*\sin(ωt+Φ)$$ then it is given that potential drop across the capacitor is V.ie$$V=-iωC*\cos(ωt+Φ)$$ but according to my derivation answer comes something diffrent. Note C is capacitance of capacitor The derivation is as follows:-

$$I=dq/dt$$ dq/dt=rate at which charge increases on plate of capacitor Note q is charge on plate of capacitor $$∫dq=∫Idt=∫isin(ωt+Φ)dt$$

$$q=-(icos(ωt+Φ))/ω + d$$

d is constant of integration

Now at t=0 the q=0 so on substituting the condition into the equationwe get d and our final equation is

$$q=(i(cos(Φ)-cos(ωt+Φ))/ω$$

hence the potential diffrence across capacitor is $$V=(i(cos(Φ)-cos(ωt+Φ))/Cω$$

why the extra term appears in derivation.How to eliminate this term

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  • $\begingroup$ Shouldn't the 3rd equation be $V = -\frac{i}{\omega C}\cos(\omega t + \Phi) + \bar{V}$ where $\bar{V}$ is the mean voltage across the capacitor (which is typically zero)? Also, I recommend changing the notation to something more standard, e.g., $i(t) = I_0\sin(\omega t + \phi)$ $\endgroup$ – Alfred Centauri Aug 14 '19 at 18:28
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    $\begingroup$ Make $\Phi=\pi/2$? $\endgroup$ – BioPhysicist Aug 15 '19 at 3:28
  • $\begingroup$ @AlfredCentauri reactance of a capacitor is $1/(\omega C)$ $\endgroup$ – Anonymous Nov 18 at 6:16
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Well, we know that the complex voltage across the capacitor is given by:

$$\underline{\text{V}}_{\space\text{C}}=\underline{\text{I}}_{\space\text{C}}\cdot\underline{\text{Z}}_{\space\text{C}}=\underline{\text{I}}_{\space\text{in}}\cdot\frac{1}{\text{j}\omega\text{C}}=\frac{\underline{\text{V}}_{\space\text{in}}}{\underline{\text{Z}}_{\space\text{in}}}\cdot\frac{1}{\text{j}\omega\text{C}}\tag1$$

The input impedance is given by:

$$\underline{\text{Z}}_{\space\text{in}}=\text{R}+\frac{1}{\text{j}\omega\text{C}}+\text{j}\omega\text{L}\tag2$$

And the complex input voltage is given by:

$$\underline{\text{V}}_{\space\text{in}}=\hat{\text{u}}\cdot\exp\left(-\frac{\pi}{2}\cdot i\right)\tag3$$

So:

$$\underline{\text{V}}_{\space\text{C}}=\frac{\hat{\text{u}}\cdot\exp\left(-\frac{\pi}{2}\cdot i\right)}{\text{R}+\frac{1}{\text{j}\omega\text{C}}+\text{j}\omega\text{L}}\cdot\frac{1}{\text{j}\omega\text{C}}=\frac{\hat{\text{u}}\cdot\exp\left(-\frac{\pi}{2}\cdot i\right)}{1-\omega^2\text{CL}+\text{R}\omega\text{C}\text{j}}\tag4$$

And, so the time-expression is given by:

$$\text{V}_{\space\text{C}}\left(t\right)=\left|\underline{\text{V}}_{\space\text{C}}\right|\cdot\cos\left(\omega t+\arg\left(\underline{\text{V}}_{\space\text{C}}\right)\right)=$$ $$\frac{\hat{\text{u}}}{\sqrt{\left(1-\omega^2\text{CL}\right)+\left(\text{R}\omega\text{C}\right)^2}}\cdot\cos\left(\omega t-\frac{\pi}{2}-\arg\left(1-\omega^2\text{CL}+\text{R}\omega\text{C}\text{j}\right)\right)\tag5$$

Now, let's assume that $1-\omega^2\text{CL}>0$, then we know:

$$\text{V}_{\space\text{C}}\left(t\right)=\frac{\hat{\text{u}}}{\sqrt{\left(1-\omega^2\text{CL}\right)+\left(\text{R}\omega\text{C}\right)^2}}\cdot\cos\left(\omega t-\frac{\pi}{2}-\arctan\left(\frac{\text{R}\omega\text{C}}{1-\omega^2\text{CL}}\right)\right)=$$ $$\frac{\hat{\text{u}}}{\sqrt{\left(1-\omega^2\text{CL}\right)+\left(\text{R}\omega\text{C}\right)^2}}\cdot\sin\left(\omega t-\arctan\left(\frac{\text{R}\omega\text{C}}{1-\omega^2\text{CL}}\right)\right)\tag6$$

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  • $\begingroup$ It might be painful to type this level of mathjax $\endgroup$ – Anonymous Nov 18 at 6:16

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