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How does the LC circuit respond to the DC input?

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Initially, when the switch is closed, at that instant the inductor is a short and capacitor is open so the diode cathode is floating. How does the diode conduct?

Does the LC circuit resonate?

My attempt at the solution:

Under steady state the capacitor should charge to 100V and inductor voltage approaches 0V, so the answer should be 100 volts

The ambiguity regarding resonance came in because of the natural response had sinusoids in it.

The answer is 200 volts, but I don't understand it.

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    \$\begingroup\$ Homework questions with no attempt at a solution are closed. \$\endgroup\$ Oct 2, 2015 at 12:24
  • \$\begingroup\$ "Does the LC circuit resonate?" What do you think? \$\endgroup\$ Oct 2, 2015 at 12:26
  • \$\begingroup\$ the confusion is that when i solved for the natural responce in the absence of forcing input ,i got sinusoidal solution with natural frequency of 1/sqr_root (LC) \$\endgroup\$ Oct 2, 2015 at 12:30
  • \$\begingroup\$ @BrianDrummond i am confuse regarding that ,however i have updated my approach,pls correct me if i am wrong \$\endgroup\$ Oct 2, 2015 at 12:41
  • \$\begingroup\$ @LeonHeller this aint any homework question, i have been solving many interesting questions and when i am in doubt with my attempt at the solution ,i try to clear it out with you all.Also i have updated my attempt on the question!! \$\endgroup\$ Oct 2, 2015 at 12:54

3 Answers 3

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200 V is indeed the correct answer, assuming ideal components.

Think about this in the time domain. When the switch is initially closed, all the 100 V of the battery is applied across the inductor. Voltage across a inductor causes current thru it to rise linearly, with the slope proportional to the voltage. Immediately after turnon, the inductor current will therefore rise linearly.

However, the capacitor voltage builds up with the integral of this current. After a little while, the capacitor will have charged up a little, and the voltage across the inductor therefore reduced. This reduces the rate of current increase, but note that the current is still increasing. This current causes more voltage to build up across the cap, which decreases the voltage on the inductor, which decreases the rate of rise of the current.

Eventually the cap voltage builds up to the same as the supply voltage. At that point the voltage on the inductor is zero. However, that only means the current stops increasing, not that it stops. In fact, this is the point with the largest current.

Since the current keeps flowing, voltage on the cap keeps building up, which is now so high that the voltage across the inductor is negative, and the current starts going down. Eventually, this negative voltage on the inductor brings the current to zero.

However, at that point, the cap has charged up to twice the supply voltage. If the diode weren't there, the reverse voltage on the inductor would continue decreasing the current, now making it negative. This would eventually discharge the cap until it is at zero. That causes the current to increase, and the whole thing happens again. With ideal components, both the current across the inductor and voltage across the capacitor are sines, with the system oscillating continually ad infinitum.

In your case, the diode prevents the current from going negative. The system stops changing when the zero current point is reached, then stays that way forever (again, with ideal components). At that point the cap has 200 V on it.

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  1. Your first statement is wrong. After closing the switch, L is by no means a short. Look at the properties of an inductor. U is proportional to the first derivative of I. Hence, I must be differentiable all the time. A short allows instant changes of I which makes it discontinuous.

  2. Solving the problem in the frequency domain by calculating the resonance frequency is a bad idea, as the situation deals with step responses and nonlinear elements.

  3. The same false claim from 1) fools you about the behaviour of the circuit when C has reached a voltage of 100V. You have to take into account, where the magnetic energy is going after the circuit settles.

In fact, after C has reached 100V there's further current through L, driven by the magnetic field in L.

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Here are the formulas to consider: $$i = Cv^{'}_C \,where \,v \,is \,the \,potential \,difference$$ $$v^{''}_C + w^{2}_0v^{}_C = w^{2}_0v^{}_i$$ $$v^{}_C(t) = Asin(w^{}_0t) + Bcos(w^{}_0t) + v^{}_i$$ $$v^{}_C(0) = 0 = B + v^{}_i \Longrightarrow B = - v{}_i$$ $$v^{}_C(t) = Asin(w^{}_0t) + v^{}_i(1 - cos(w^{}_0t))$$ $$i(0) = 0 \leftrightarrow v^{'}_C(0) = 0 \leftrightarrow A = 0$$ $$v^{}_C(t) = V^{}_i(1 - cos(w^{}_0t)) = 2V^{}_isin^2\left(\frac{w^{}_0t}{2}\right)$$ $$w^{2}_0 = \frac{1}{LC}$$ $$T = \frac{2\pi}{w^{}_0}$$ $$v^{}_L = L^{'}_i = LCv^{''}_C = \frac{v^{''}_C}{w^{2}_0} = V^{}_icos(w^{}_0t)$$

The steps are: $$V^{}_i = 100 V$$ $$V^{}_{C^{}_{MAX}} \,increases \,to \, t \,at \,the \,maximum \,value \,of \,200 \,V$$ In reality $$V^{}_{C^{}_{MAX}} = 200 - V^{}_{DIODE} \lesssim 200 \,V$$ At this instant (about T/2) the diode is off and the capacitor stays charged at about 200 V

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