Why Does Doubling the Time Solve the AC Transient Equation in an RC Circuit?

In summary, In this RC circuit, the voltage source should be connected to the circuit at its zero crossing.
  • #1
cnh1995
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Homework Statement
To find the instant at which switch is closed for transient-free response.
Relevant Equations
For transient free response in an RC series circuit, the voltage source should be connected to the circuit at its zero crossing.
20200615_195143.jpg


Here, at voltage source zero crossing,
cos(2to + pi/4) =0.
So, 2to+pi/4 = pi/2
This gives to=pi/8 = 0.3926s.

But the given answer is twice the answer I got i.e. 0.7852s.

Have I missed anything here?
 
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  • #2
I get the same answer as you. Maybe do a quick Excel spreadsheet to plot the function to see if we're missing something...
 
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  • #3
berkeman said:
I get the same answer as you. Maybe do a quick Excel spreadsheet to plot the function to see if we're missing something...
The book is not much reliable in terms of the answer keys, it's just a good collection of problems. I ran a quick simulation on my phone and the result is in agreement with my answer.
 
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  • #4
Sigh... Another poorly worded homework problem.

This isn't solvable without knowing the initial condition of the circuit; in this case, the capacitor voltage at t0. However, it is reasonable for you to assume it's zero. His answer is correct if the initial voltage is -7.07V.

You get an "A", your instructor gets an "F".
 
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  • #5
DaveE said:
Sigh... Another poorly worded homework problem.

This isn't solvable without knowing the initial condition of the circuit; in this case, the capacitor voltage at t0. However, it is reasonable for you to assume it's zero. His answer is correct if the initial voltage is -7.07V.

You get an "A", your instructor gets an "F".
I went through their solution.
For an RC circuit with sine and cosine inputs, there are two standard formulae for transient-free condition. In this problem, as the input is cosine, the formula for cosine input gives to=0.7852s.
I need to study their derivations and then compare them with my logic.

Never imagined RC transients will be this much tricky.
I need to stop underestimating concepts..o0)
 
  • #6
cnh1995 said:
For transient free response in an RC series circuit, the voltage source should be connected to the circuit at its zero crossing.
This is incorrect. It is true only for a purely capacitive circuit. In my simulation, my values for R and C might have been small, so I may have misread the output transient waveform. o0)

After analysing this using differential equations, it turns out that the source should be connected at the "steady state current-peak" angle i.e. at the phase angle when the steady state current will peak.
In this problem, the steady state current leads the input voltage by pi/4 rad. So, the steady state current will be I(t)=I*cos(2t+pi/4+pi/4).
It peaks when cos(2t+pi/4+pi/4)= +1 or -1.

This indeed gives to=pi/4=0.7852s.
 
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  • #7
cnh1995 said:
This is incorrect. It is true only for a purely capacitive circuit. In my simulation, my values for R and C might have been small, so I may have misread the output transient waveform. o0)

After analysing this using differential equations, it turns out that the source should be connected at the "steady state current-peak" angle i.e. at the phase angle when the steady state current will peak.
In this problem, the steady state current leads the input voltage by pi/4 rad. So, the steady state current will be I(t)=I*cos(2t+pi/4+pi/4).
It peaks when cos(2t+pi/4+pi/4)= +1 or -1.

This indeed gives to=pi/4=0.7852s.

Are you still discussing the circuit you originally posted? Because now I am completely confused. Before t0, there is no current flow. When the switch is closed, the current is solely determined by the voltage difference across the resistor. Since it was 0, for no current transient, it should remain 0 when the switch is closed, which means the source voltage should initially match the capacitor voltage, right?

cnh1995 said:
Never imagined RC transients will be this much tricky.

I actually think this is one of the simplest transient analysis problems you are likely to ever see.

BTW, there is no steady state as described. This is a transient problem.
 
  • #8
DaveE said:
Are you still discussing the circuit you originally posted?
Yes.
The capacitor was uncharged initially (a valid assumption).

I agree the problem is poorly worded.
A more accurate problem description would be somewhat like this:
The switch is open for a long time and the capacitor is initially uncharged. If the switch is closed at t=to and the observed response is transient-free,
find to.

This is a problem asking for the switching phase angle of the voltage source which eliminates the dc transients in this ac circuit. In an RL circuit, this switching phase angle is equal to the power-factor angle i.e. the phase angle that represents the 'zero crossing' of the steady-state current.

In an RC circuit such as this one, this switching phase angle is 90-power-factor angle i.e. the phase angle that represents the 'peak' of the steady-state current.
 
  • #9
Sorry my reply sounded a lot more rude than I intended. What I meant was that you are over-thinking the original problem.

For your new problem, I think I need to see a schematic or a detailed description of the control law for the switch. It sounds like a standard SCR power supply problem where the switch is synchronous with the source frequency and the output is controlled by the phase angle when the switch turns on? Or maybe a ZCS SMPS problem? This new problem is definitely not simple.

BTW, I am at a complete loss in understanding "DC transient". This is an oxymoron in my world. I guess you mean the low frequency components of the transient response? Which begs the question "what changed" to make the "DC" change? In general, please be clear about the nature of the problem and solution you seek. There are typically DC, AC steady state, transient, and repetitive transient problems, and usually the analysis is best done by focusing on one at a time. This is the beauty of linear systems, superposition usually allows you to break up the analysis.

Anyway, it sounds like you know the "tools" to solve these sort of problems; i.e. matching initial conditions in transient solutions. But making the question clear, either in your mind or in mine (I'm not sure which right now) is the issue.
 
  • #10
Final Edit: the given answer is correct!

Just an outline:

The criterion for "zero transients" must be current continuity with only sines and/or cosines but no net exponential decay term in the current i(t).

The generated excitation voltage is ## v_g(t) = cos(\omega t + \pi/4) ##.
The switch modifies the excitation to
## v(t) = v_g(t) U(t-T) ##
where T is the moment of switching ON.
## L[v_g(t) U(t-T)] = e^{-sT} L[v_g(t+T)] ##
where L signifies Laplace transformtion.

I expanded v_g into sine & cosine, and then also ## L[v_g(t+T)] ## into sines and cosines in t and T. Then I laplace transformed with respect to t.

At some point I came up with having to inverse-transform
## \frac {s^2} {(s^2+\omega^2) (sRC+1) } ##

and also

## \frac {\omega~s} {(s^2+\omega^2) (sRC+1)} ##

both of which I reduced in partial fractions to 2 terms each.

It was immediately obvious that the first term in each partial fraction pair
transforms to pure sines and cosines. No problems there.

But the second term in each pair transformed to an exponent ## e^{-t/RC} ##.

You then also need ## e^{-sT} F(s) \leftrightarrow f(t-T) U(t-T) ## if
## F(s) \leftrightarrow f(t) ##.

I wound up with
## sin(\omega T)~(1/R -\omega C) = cos(\omega T) (1/R + \omega C ) ## or

## \omega T = tan^{-1} (1/R + \omega C)/(1/R - \omega C) ##.
For our problem, ## \omega = 2, R = 1, C = 1/2 ##, so ##\omega T = \pi/2 ##

## T = \pi / 2 \omega = \pi/4 = 0.78854 s. ##
 
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Related to Why Does Doubling the Time Solve the AC Transient Equation in an RC Circuit?

1. What is an AC transient in an RC circuit?

An AC transient in an RC circuit refers to the temporary change in voltage or current that occurs when an alternating current (AC) is applied to a resistor-capacitor (RC) circuit. This transient response is caused by the charging and discharging of the capacitor, which creates a time-varying voltage or current.

2. How does an RC circuit behave during an AC transient?

During an AC transient, the capacitor in an RC circuit initially acts as a short circuit, allowing the current to flow through it. As the capacitor charges, the current decreases and the voltage across the capacitor increases. Once the capacitor is fully charged, it acts as an open circuit, stopping the flow of current. As the AC signal changes direction, the capacitor discharges and the cycle repeats.

3. What is the time constant of an RC circuit?

The time constant of an RC circuit is a measure of how quickly the capacitor charges or discharges. It is equal to the product of the resistance (R) and the capacitance (C) in the circuit. A larger time constant means the capacitor takes longer to charge and discharge, resulting in a slower transient response.

4. How does the frequency of the AC signal affect the transient response in an RC circuit?

The frequency of the AC signal has a significant impact on the transient response in an RC circuit. A higher frequency means the AC signal changes direction more frequently, resulting in a faster charging and discharging of the capacitor. This leads to a shorter time constant and a more rapid transient response.

5. What is the significance of AC transients in RC circuits?

AC transients in RC circuits are important to understand because they can affect the overall performance and stability of the circuit. They can also cause distortion in the output signal and affect the accuracy of measurements. By analyzing the transient response, scientists and engineers can design and optimize RC circuits for specific applications.

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