Parallel RC CIrcuit Using Thevenin Equivalent
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- เผยแพร่เมื่อ 24 ม.ค. 2018
- Physics Ninja shows you how to solve a Parallel RC circuit using a Thevenin Equivalent Network. I previously solved this circuit using Kirchhoff's Laws.( • Parallel RC circuit )
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Thanks for your video! This made figuring out one of the questions on my circuits final exam a lot easier.
Great video, with simple and clear explanation. Thank you!
Thank you sooo much sir, you've just saved my homework
Thanks for giving a short and brief solution. I was banging my head against the wall trying to model a lithium ion battery equivalent RC circuit using Laplace transform!
Glad it helped
Very helpful, thank you so much!
Wonderful! I might add that source transformation works a bit easier in this kind of problem!
Thank you!! Great explanation....
I watched both videos and agree the Thevenin approach is so much easier. Prof never mentioned it in my physics class!
It's rarely covered in physics classes, you may come across it in a 2nd year electronics class. I just learned the technique a few month ago!
@@PhysicsNinja In India they teach us this for IIT JEE, in class 12
Very helpful lecture thanks friend
earned a sub
Thanks a lot man!!!
I was struggling with this problem whole day and asked about and nobody could help me, then i found your video using kirchoff law and got even more confused as i couldnt get how to calculate the voltage using all that 30min of calculation
And now this is like a minute of calculation and done!
Thank you so much...
Thevenin equivalent allow one to solve very complex circuits in minutes. Kirchoff's will always work but takes more time.
Thanks a lot, sir!
Good lecture..
Plz show some more complicated circuits
tq this help me alot
u r the best
Nice lecture
Damn good video.
nice video and explanation..but you should have taken another example solving a little more complex question so that we could get a good idea of solving such problems
Sehr gut
When you want to find charge over time and max charge etc. do you plug in the thevenin voltage into Q=CV. I thought you would plug in the emf of the battery as the capacitor would take all the voltage but now I'm not really sure.
if you have a resistor in series with the capacitor, would you solve it exactly the same except the time constant would be: TC = (Rth + R[series] ) C
Yes! Thevenin equivalent can be used to simplify the network and you'll have the Rth in series with the other resistor and the capacitor.
If I have V as a sinus function, (Vpp) do i transform my Vpp to Vrms? or do i run with the Vpp value?
Great video! How would the voltage response look like if a constant current was forced through the circuit (instead of applying a constant voltage)?
ignore R1 since the current going into R1 is the same as going into R2 and C. Perform a current to voltage source transformation by putting R2 and C in series with a voltage source equal to R2*I. Use voltage equation for capacitor; V(t)=V(inf)+(v(0)-v(inf)*e^-t/tau. V(inf) is equal to R2*I since this is the source voltage and v(0)=0 since initial capacitor voltage is zero. subbing in values this gives v(t)=R2*I-R2*I*e^-t/R2*C
you could also do a Norton equivalent. Rn=Rth and to find the theneviven resistance across the capacitor you open circuit your current source meaning that Rth=R2 since it is the only resistor that current can now travel across. IN=I since the current going through R2 is the same as R1 as shown previously resulting in R1 being ignored. You perform a source transformation as previously stated and end up with the same thenevien voltage equivalent circuit with the same solution and answer.
Can you tell after how many time constants the current through the resistor R2 is equal to the current through the capacitor or when the voltage drop over the capacitor is equal to the voltage drop over R2?
You said at 5:55 that you short out the batteries when you're finding the thevenin resistance. If you have a current source instead, do you also short it out or do you have a break?
Also thanks for making such a clear video 👍
James Morley check out my other video comparing thevenin to norton.
Open circuit for current sources ( i=0)
Conceptually, I was wondering if we swapped R2 and C, would the problem remain the same. I believe that it would as the current is still split between the two parallel branches, but if am unsure if the resistors being directly in line with each-other has an effect that I am overlooking. Thanks!
No difference if you flip R2 and C. I have another video where I solve the same problem without thevenin.
I'm actually wondering why we need R1. When I put up the circuit in a simulator without R1, it says there is no current whatsoever.
What would happen if you added another resistor in the right loop, between the capacitor and the next junction?
you can still find a thenevin equivalent
Then the resistor is in series so just add it with Rth when calculating time constant
What SW is he using to draw while he is explaining?
Mario Arenas I use Notability on Mac with a Wacom tablet
Physics Ninja Nice break down of the Thevenin theorem
Solve for Vth
Voc= Vr2= V+Vr1
i*r2= V+ i*r1
i= V/(r2- r1)
Vth= r2*V/(r2-r1)
what am i doing wrong here? cause i am getting different value for the current
never mind! V= i*r1+i*r2 because of the voltage drop across the resistances, my bad..
This looks wrong. You claim R1 and R2 are in parallel but add the resistances at 5:29 to make an equivalent resistance (for series connections)
How will we calculate if a capacitor is connected in series with R1
just use the equations he used for the thenevin circuit since that is just a resistor and capacitor in series.