Waveform of R,L,C in DC circuit

[=SJ=]

Homework Statement

Draw a time-depended waveform of all components in DC RLC circuit.
R=4*1,412
L=1,6 mH
C=0,2 mF
Udc=15 V

Homework Equations

The Attempt at a Solution

DoCircuit simulator

I am a bit confussed with this homework because I am aware that capacitor do not conduct a current in DC circuit.
Can somebody help me meet my homework, please?

 

[LvW]

Your assumption (no DC current within a capacitor) is valid under steady-state conditioins only.
Which means: All switch-on transients have disappeared (theoretically for time t approaching infinite).
Thus, the task is to calculate the timely behaviour of the current directly after voltage is applied to the circuitry (inrush current).
Therefore, you have to combine the current-voltage relationships for all three components for NON-SINUSOIDAL waveforms (that means: NOT V=I*Z).
Example: V(L)=L*(dI/dt).

 

[=SJ=]

Thank you very much for reply.
I am not sure how perform calculation or simulation of such a circuit.
I have here a sim, which can show AC Transient parameters but I do not know how to apply it for DC.

 

[gneill]

Thank you very much for reply.
I am not sure how perform calculation or simulation of such a circuit.
I have here a sim, which can show AC Transient parameters but I do not know how to apply it for DC.

For the simulation you would need to replace the AC voltage source with a DC source that is switched on at time t = 0. The software probably has a selection for a "pulse" waveform or the ability to set the sources to start from 0V at "turn-on". You would also have to replace the voltmeters with oscilloscopes so that you can "see" the resulting waveforms.

For calculation you need to analyze the circuit. It's behavior is governed by a differential equation which you can write using the circuit laws (KVL, KCL). If you are not required to show that derivation then you can look up the solution online ("RLC circuit").

 

[=SJ=]

I was not able at last conduct any calculation. Actually, I don't even know how to connect the oscilloscope property.
Any help would be much appriciate. Thanks.

 

[gneill]

Going by the image, the oscilloscope has two separate channels. As you've drawn it currently you have the channels connected to either end of the resistor. It would appear that each channel will measure a signal with respect to ground reference. So if you have a ground connection specified on your circuit (as you have in the diagram connected to the negative terminal of the battery), then you need only a single wire from a channel connector to a point on the circuit. The oscilloscope should then be able to display the voltage at that point of the circuit with respect to ground. As you've drawn it currently you have the channels connected to either end of the resistor.

Try moving the ground around in order to place the oscilloscope's reference point at different locations. So for example you might place the ground at one end of the resistor and a scope channel at the other end, thus displaying the voltage across the resistor.

I should mention that I am totally unfamiliar with this DoCircuit software, so I won't be of much help with details! When I want to do circuit simulations I generally choose LTSpice.

 

[LvW]

I was not able at last conduct any calculation.

I suppose you are able to write down the three voltages created across the tree elements R, L and C.
In my post#2 I gave you already the voltage across the L element.
The sum of these three voltages equals the driving DC voltage, OK?
Now you have an equation like
Udc=V(R)+V(L)+V(C).
This equation can be transferred into a second order differential equation (differentiate the whole equation).
This diff. equation can be solved setting i(t)=I*exp(st).
At the end of the calculation you have to determine two constant quantities (currents) using the initial conditions at t=0.

 

[=SJ=]

I haven't manage to remove grounds of the oscilloscope in DoCuircuit, so I have created the circuit in LTSpice and watched some tutorials on YouTube. Therefore it still doesn't work for me.

Yes, I know these equtions, actually I have calculed a critical resistor value using them (attached). But I am not sure how to applied them in accordence to obtain graphs. We can't be ask to make an analysis of them and draw the charts by hand (there's a lot more of them), can we?

 

[amilapsn]

what is the value of R in ohms?

 

[LvW]

=SJ=, I cannot read your png attachement. Bad quality, please improve the contrast.

 

[=SJ=]

The resistor value is 5.656 ohms.
Actually, the paper does not matter so much, I am chiefly intersted in simulations to plot graphs.
Any ideas?

 

[LvW]

The resistor value is 5.656 ohms.
Actually, the paper does not matter so much, I am chiefly intersted in simulations to plot graphs.
Any ideas?

The task is to plot "time-depended waveform of all components".
What does this mean (waveform of all components)? Most probably: Voltages across all three components.
As mentioned already - excite the series combination of the three componenets with a dc voltage that is switched on at t=0.
This requires neither an ac analysis nor a dc analysis but a TRANSIENT analysis which reveals the time-dependent behaviour of the circuit.
Is your simulation program able to perform such an analysis?

 

[NascentOxygen]

DoCircuit simulator

I am a bit confussed with this homework because I am aware that capacitor do not conduct a current in DC circuit.
Can somebody help me meet my homework, please?

Hi SJ. Your best resource for all of this is your fellow student. You should be discussing homework requirements and software details with other students studying the subject with you. Not copying their work, but collaborating and helping each other to understand these topics. If you try and work in isolation you will find some subjects are a struggle and not achieve to your potential. A few minutes' discussion with another in your class can save you hours of individual toil and worry. Make some friends quickly! You won't be alone in experiencing difficulty understanding some of this.

 

[=SJ=]

LvW:
- I measure voltage across all three components
- It is powered by dc voltage at t=0
- I have been always performing TRANSIENT analysis

The only issue which I am facing is that, in both DoCircuit and LTSpice such an analysis of DC powered RLC shows nothing and I do not why.

 

[=SJ=]

NascentOxygen:
You are absolutely right but unfortunately I study in combined program (work/school), have moved this subject from previous semester and missed the only one lecture of this subject for combined students (just a few of us). It seems I have to done the work entirely by myself and I feel a solution can't be far away.

 

[LvW]

The only issue which I am facing is that, in both DoCircuit and LTSpice such an analysis of DC powered RLC shows nothing and I do not why.

I cannot help you, unless you are a bit more specific.
What means "shows nothing"? A straight line or a black screen or something else?
You should show us your LTSpice simulation arrangement (circuit and voltages vs. time).

 

[=SJ=]

Here's printscren. Please, let me know if any futher informations are required.

 

[=SJ=]

I have just notice I didn't set "start at 0V" in LTSpice.
Looks good now for me. What are saying about?

 

[LvW]

I have just notice I didn't set "start at 0V" in LTSpice.
Looks good now for me. What are saying about?

Yes - everything seems to be OK. Because of the large resistor value (1k) the system is overdamped -no oscillations.
Try a smaller value (100ohms, 10 ohms) - and you will see some damped oscillations.

 

[NascentOxygen]

Looks very promising!

 

[=SJ=]

Thank you very much!

 

[=SJ=]

One futher question - how can it be, the capator reachs in peak up to 2V while the voltage source has 1V?

 

[NascentOxygen]

The capacitor voltage rises towards 1v but overshoots and reaches 2v, before falling back towards 0. That's what your graph shows? Sounds right.

 

[LvW]

One futher question - how can it be, the capator reachs in peak up to 2V while the voltage source has 1V?

Yes -that is a "normal" effect for oscillating circuits. The "key word" behind this effect is "phase shift".
Remember that - at each moment - the sum off all voltages across the three elements must be identical to the driving voltage (KVL)
Now - because the voltages across the inductor and across the capacitor are out of phase by 180deg , both amplitudes are substracted from each other.
As a consequence, one must be larger than 1V. The remaining voltage (difference) - added to the voltage across the resistor - is identical to the driving voltage.
This can be seen in the graph of all three voltage drops: At each time slot the sum of all voltages is identical to 1V (DC input).

 

[=SJ=]

Can see now.
Thank you very much again!