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Suppose the circuit is in steady state when the switch is at position (1). At t=0 the switch goes from position (1) to position (2)and the problem is to find the response of the system. I am having some difficulties finding the initial value for the voltage on the capacitor. Since it was stated that we can assume it is in steady state for t<0, I solved it in the phasor domain and got that the rms value of Uc=7.01V. Can I somehow use this for the actual initial value(in the time domain). Data given:

  • R = 10,
  • L=10 mH,
  • C=100 uF,
  • E = 10V,
  • e(t)=14.1*sin(1000t+ 135(deg)).

EDIT I see a lot of arguments in the comments below, and one question that basically repeats what is said in the question. I would be very grateful if someone could give a concise answer, how to solve it, and if it is not possible, why it is not possible. EDIT 2 Thanks to everyone who was active in the comments section, and who answered. I will award the best answer. Basically, the easiest way is to solve the circuit in the phasor domain, reconstruct the equations for the time domain, and just plug in the value for t=0, since we are viewing the system for t=\$-\infty\$ to t=\$\infty\$

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  • \$\begingroup\$ The initial voltage of the capacitor C depends on the function of the voltage source e(t). Please specify it in your question. \$\endgroup\$ Nov 21 '15 at 12:40
  • \$\begingroup\$ @MartinZabel it is specified. \$\endgroup\$
    – Andy aka
    Nov 21 '15 at 12:41
  • \$\begingroup\$ @MartinZabel I did \$\endgroup\$ Nov 21 '15 at 12:41
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    \$\begingroup\$ The circuit has attained sinusoidal steady state before switching. Substituting t=0 in sine function would mean that the transients have not died down. You can't find the instantaneous value of capacitor at the moment of switching unless you know the value of variable 't' of sine function. \$\endgroup\$
    – JGalt
    Nov 21 '15 at 14:57
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    \$\begingroup\$ @Aditya Patil, no, it means that the sinusoid has been going since \$t=-\infty\$, then at t=0 the switch is operated. \$\endgroup\$
    – Chu
    Nov 21 '15 at 16:57
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I would be very grateful if someone could give a concise answer, how to solve it, and if it is not possible, why it is not possible.

It is possible. For \$t \lt 0\$, the switch is in position 1 and you have a series RC circuit in sinusoidal steady state.

The phasor voltage across the capacitor in a series RC circuit is given by:

$$V_c(\omega) = V_s\frac{1}{1 + j\omega RC} = V_s \frac{e^{i\phi}}{\sqrt{1 + (\omega RC)^2}}$$

where \$\tan \phi = -\omega RC\$.

In the case of the circuit in your question

$$V_s = 10\sqrt{2}\;e^{i\frac{\pi}{4}}$$

$$\omega RC = 1$$

Before the switch changes state, the RC circuit is in steady state and so

$$V_c = 10\sqrt{2}\;e^{i\frac{\pi}{4}}\frac{e^{-i\frac{\pi}{4}}}{\sqrt{1 + 1}} = 10\mathrm V$$

Converting to the time domain, the capacitor voltage is then

$$v_C(t) = 10\mathrm V \cos(1000\cdot t)\;, t \le 0$$

and thus the capacitor voltage, just before the switch changes state is

$$v_C(0-) = 10 \cos(0-) = 10 \mathrm V $$

Since the voltage across a capacitor cannot instantaneously change (for finite current through), the voltage across the capacitor just after the switch changes state is unchanged.

$$v_C(0+) = v_C(0-) = 10\mathrm V $$

This is the initial capacitor voltage that is required to find the solution for \$t \gt 0\$

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  • \$\begingroup\$ Isn't the rms value actually 10/sqrt(2)? By defintion it should be less than the amplitude in the t-domain- \$\endgroup\$ Nov 22 '15 at 8:21
  • \$\begingroup\$ @Shemafied, it's true that the rms value of \$10 \cos (1000t)\$ is \$10 / \sqrt{2} \$ but why do you want the rms value of the capacitor voltage? By the way, and if it isn't clear, I'm using peak phasors in my calculation rather than rms phasors \$\endgroup\$ Nov 22 '15 at 14:05
  • \$\begingroup\$ That's what caused the confusion. \$\endgroup\$ Nov 22 '15 at 19:24
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Firstly, note that \$\small 14.1\$ is probably \$\small 10\sqrt 2\$.

At \$\small \omega=1000\$ rad/sec, \$\small X_C=-j10\$. Hence, the voltage across the capacitor, relative to a sinusoidal voltage source of amplitude \$\small 10\sqrt2\$, but with zero phase angle, \$\small \phi = 0^o\$, would be:

\$\small V_C=10\sqrt2\:\dfrac{-j10}{10-j10}= 10\sqrt 2\:\dfrac{-j}{1-j}=10\sqrt 2\:\dfrac{1-j}{2} \equiv 10\:\large\angle\small{-45^o}\$.

However, the source actually has a phase angle, \$\small \phi=135^o\$, hence the actual \$\small V_C\$ is:

\$\small V_C=10\:\large\angle\small (135-45)=10\:\large \angle\small {90^o}\$.

Thus, when \$\small t=0\$, the capacitor voltage is: \$\small V_C=10\:sin(90^o)=10V\$, which is the required initial condition.

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It is not enough to know the RMS (or the amplitiude) of the capacitor voltage as it does not tell you the voltage exact at the moment the switch is changed (t=0) but only the average over time (or the maximum).

If you want to use phasor domain analysis to find the initial condition of the capacitor you can still use it: you must also use the phase angel. Together with the amplitude it tells you the momentary voltage at t=0.

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  • \$\begingroup\$ So you suggest to solve the circuit( pos(1)) using the classical approach, and then just use whatever values I get at t=0 as my initial conditions. Is it even possible to solve the part for (1) since we don't have the initial charge on the capacitor at t=$\-infty$, or do we just assume it is zero? \$\endgroup\$ Nov 21 '15 at 14:32
  • \$\begingroup\$ @Shemafied "and then just use whatever values I get at t=0 as my initial conditions." No. t=0 is just a way of representing state of the circuit at the moment of switching. It has nothing to do with the sinusoidal function. \$\endgroup\$
    – JGalt
    Nov 21 '15 at 14:44
  • \$\begingroup\$ @Aditya Patil, no, you're wrong, the 't' in the sinusoid is the same 't' as in the initial condition, t=0. \$\endgroup\$
    – Chu
    Nov 21 '15 at 16:52
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    \$\begingroup\$ There is only one letter t in the equations! We have two expressions: \$t=0\$ and \$e(t)=14.1 sin(1000t +135^o)\$, why would you think they refer to different times? Apart from anything else, the question wouldn't make sense with your interpretation. \$\endgroup\$
    – Chu
    Nov 21 '15 at 18:16
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    \$\begingroup\$ @AdityaPatil, it doesn't need to be said since it is understood that \$t\$ is the time parameter by almost everyone familiar with these types of problems. The source voltage is specified as a sinusoidal function of the time parameter \$t\$ where \$-\infty \lt t \lt \infty\$. For this problem, the switch has been in position 1 for all time before \$t=0\$ and is in position 2 for all time after \$t=0\$. \$\endgroup\$ Nov 21 '15 at 21:50
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EDIT: according to the discussion in the comments, the following applies only for DC steady state.

For me a steady state means that the properties of a system do not change there values with time, i.e, for a property p $$dp/dt=0$$

Thus voltages, currents and the electric charge whithin the capacitance need to be static.

This rule is violated here because the source e(t) tries to change the voltage for t<0. Thus, there is no steady state while the switch is in position 1 as claimed in the question.

(Hence why I asked in the comments below the question.)

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  • \$\begingroup\$ t can be negative! \$\endgroup\$
    – Chu
    Nov 21 '15 at 17:04
  • \$\begingroup\$ @Chu Yes of course. It has no impact on my answer. Point t=0 is just the moment when the switch is changed to position 2. \$\endgroup\$ Nov 21 '15 at 17:07
  • \$\begingroup\$ it's also the time when the sinusoid is \$14.1sin(0 +135^o)\$. There is only one time, t, in the question. \$\endgroup\$
    – Chu
    Nov 21 '15 at 17:46
  • \$\begingroup\$ No, the circuit is operating in sinusoidal steady state. Steady state does not strictly mean 'not changing with time'. The steady state here refers to the state attained after transients have died down. \$\endgroup\$
    – JGalt
    Nov 21 '15 at 17:56
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    \$\begingroup\$ Steady state means the sinusoid has been established for a long time and the currents and voltages in the circuit have reached their respective constant amplitudes and phase angles. It does not mean the circuit is at DC steady state. Look at the circuit diagram - with the switch in position 1 the sine voltage is applied to the R/C series circuit for a long time. You are misinterpreting 'steady state' when applied to AC circuits. \$\endgroup\$
    – Chu
    Nov 21 '15 at 18:08

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