# Inductor and Capacitor in Parallel

I've been staring at this for 10 minutes and I'm soooo confused. Hopefully, you guys can help!

So, let's say we close the switch. From what I understand, the capacitor will charge up to $$10V$$ almost instantaneously, while no current will flow through the inductor. At the very beginning, there'll be $$0.2A$$ flowing through the resistor.

But what happens if we leave the switch closed for a long time?

My book says the current through the inductor would pick up to $$0.2A$$, while the current through the capacitor drops to $$0A$$.

That first part makes sense.

However, the book then says that after an infinite amount of time, the capacitor will be at 10 volts.

That seems wrong to me...if the current through the inductor ever stops changing, the inductor would just behave like a shorted section of a circuit, right? Wouldn't the capacitor discharge through it?

In terms of voltage drop - if the current through the resistor were ever to reach $$0.2A$$, that would mean that the entirety of the voltage is dropping through the resistor. There is no voltage drop through the inductor.

Buuut, since the inductor and the capacitor are in parallel, wouldn't that automatically mean there MUST be no charge on the capacitor either?

Is my book wrong in saying after an infinite amount of time with the switch closed there will be a 10-volt difference across the capacitor?

Edit:

This is an edit after reading some of your answers and comments - they've been great. I didn't elaborate enough on what my book specifically said in my original post, so this edit is just to address that. However, the fundamentals of the question remain the same.

The exact question is this:

After a very long amount of time, the switch is opened. What is the maximum current?

Then, their explanation is as follows:

We've created an $$LC$$ circuit, and the maximum current is $$0.44A$$ - Since the current through the inductor must remain continuous, the initial current through the inductor is $$0.2A$$. The instant the $$LC$$ circuit begins to oscillate, energy is stored both within the inductor and within the capacitor. We calculate the maximum current using conservation of energy: $$\frac{LI_{max}^2}{2} = \frac{LI_{0}^2}{2} + \frac{CV_{0}^2}{2}$$. Substituting in the initial values for the current, $$0.2A$$, and for the voltage, $$10V$$, we calculate the maximum current to be $$0.44A$$.

Like I said above, that seems wrong to me because I don't think the initial capacitor voltage would be $$10V$$.

In fact, I don't think the voltage across the capacitor would EVER be $$10V$$.

$$\frac{LI_{0}^2}{2} = \frac{CV_{f}^2}{2}$$

Plugging in an initial current of $$0.2A$$, I get that the maximum voltage across the capacitor is actually: $$5.16 V$$.

• Shouldn't I = 0.2A in the beginning – Starboy Mar 31 '19 at 16:11
• @Starboy fixing it now! Thanks! – joshuaronis Mar 31 '19 at 17:33
• As for your last question, does your books perhaps say that after an infinitesimal (infinitely small, not infinitely large) amount of time with the switch closed there will be a 10-volt difference across the capacitor? – tobi_s Apr 1 '19 at 10:30
• @tobi_s Nah, and according to the currently accepted answer, even THAT would be incorrect. They say after an infinitely long amount of time - I think they just messed up. – joshuaronis Apr 1 '19 at 16:15
• Thanks, the simulation in the accepted answer simulates flipping the switch as "raise voltage of the source from 0 to 10 in 10^-6 seconds." Given that there are derivatives appearing throughout, I would assume at least some dependence of the voltages on the precise slopes. Anyway, if that's not what the book wrote, it's moot to discuss whether there are situations where this alternative version is true. – tobi_s Apr 2 '19 at 3:12

My book says the current through the inductor would pick up to 0.2A, while the current through the capacitor drops to 0A.

This is correct. To find the DC steady state solution for this circuit, replace the inductor with a (ideal) wire and replace the capacitor with an open-circuit.

Why? In DC steady state (the solution as $$t\rightarrow\infty$$), all the circuit voltages and currents are constant.

Now, recall that the voltage across an (ideal) inductor is given by

$$v_L = L\frac{di_L}{dt}$$

and so, since the inductor current is constant, the voltage across the inductor is zero. This is why you can replace the inductor with a wire.

For the (ideal) capacitor, the current through is given by

$$i_C = C \frac{dv_C}{dt}$$

and so, since the capacitor voltage is constant, the current through the capacitor is zero. This why you can replace the capacitor with an open-circuit.

In this case, it follows that both the capacitor current and voltage are zero in DC steady state.

Is my book wrong in saying after an infinite amount of time with the switch closed there will be a 10-volt difference across the capacitor?

Yes, if your book states that the capacitor has non-zero voltage across at infinite time, it is wrong for the reason I give above.

From what I understand, the capacitor will charge up to 10V almost instantaneously, while no current will flow through the inductor.

That's not correct. As the capacitor charges, the current through the inductor must increase and this inductor current means that the capacitor voltage can never reach 10V (that would require zero inductor current). This could shown by solving for the step response of the capacitor voltage which is beyond the scope of the question. However, this circuit is easily simulated with LT Spice and I've attached a plot of the capacitor voltage just after the switch is closed. See that the maximum voltage is not quite 4V.

I think your book is correct. After an infinite amount of time with the switch closed there will be no potential difference across the capacitor, there will be no current through the capacitor and there will be stationary current $$0.2$$ A through the inductor and the resistor. Do you see any contradictions in this picture?

• The thing is the book says there is a potential of 10 volts across the capacitor. In that case, wouldn't it try to discharge through the inductor? – joshuaronis Mar 31 '19 at 17:35
• @Joshua Ronis, This looks strange. The nonzero potential difference across the capacitor means that current through the inductor must change with time. – Gec Mar 31 '19 at 17:43

Good question, as the cap is charging the inductor back emf is decreasing, back emf is 0 at infinity time. Yes the cap will certainly have 10v as long as the switch is closed. The interesting part is what happens to the charge on the cap, I believe it eventually goes to zero because the bottom of the cap will be at 10v at an infinite time.

The basic rules for ideal capacitors and inductors are as follows:

Upon Switching:

You can’t change the voltage across an ideal capacitor instantaneously.

You can’t change the current through an ideal inductor instantaneously.

After a long time with the switch closed:

A capacitor will look like an open circuit

Inductor will look like a short circuit.

Alfred Centauri has shown you the equations for the basis of these rules.

So, when you say,”…the capacitor will charge up to 10V almost instantaneously” is incorrect. But you are correct that that the current through an ideal inductor cannot change instantaneously.

Now looking at your circuit diagram it appears that just prior to closing the switch, there is a charge Q on the capacitor. Then that means the voltage across the capacitor just prior to switching is

$$V_{C}=\frac{Q}{C}$$

So it appears that the inductor and capacitor are initially in parallel resonance.

Now when the switch is closed for a long time inductor is now a short-circuit with 0.2 A flowing in it and the resistor, and there is no voltage across the capacitor.

So your book is correct about the current in the inductor and wrong about the voltage across the capacitor.

Hope this helps.

• Since capacitor and inductor are parallel, shouldn't the voltage across the inductor be 10V too? That means current through the inductor is changing ( steady state has not been reached) – Starboy Apr 1 '19 at 3:06
• @Starboy See correction – Bob D Apr 1 '19 at 8:18

From my knowledge, when you close the switch, a Back-emf will be indced across the inductor which will be equal to 10 V. Now since the inductor and capacitor are connected in parallel , the Voltage across them will always be the same ( Back emf ) . Now you can apply Kirchoff's loop law in the loop containing the battery, resistor and inductor and conclude that the magnitude of back-emf decreases with time , and hence the charge on the capacitor. After a long time , the back emf across the inductor as well as the charge on the capacitor will be zero and both will behave like short-circuited paths and the current through the resistor will be 0.2A

One way to think about this is to regard the whole right hand half of the circuit diagram - both the parallel capacitor and inductor - as a single "component" (Indeed, this is just further application of the general idea of the "lumped element model" that we are using to do these kinds of simple analyses in being able to treat a circuit as composed of isolated components at all - and it's important to point out that this model has limits, however fortunately, we are well within those in this case), with its leads being the top segment and bottom segment after the resistor, in the diagram. Then you effectively have a circuit that looks like a voltage source, followed by a resistor, followed by this strange "component", in a series loop.

As a result, the total current through that "component" can never be more than $$\frac{V}{R}$$, or 0.2 A. Hence, looking at the sub-components of our "component" individually, i.e. the parallel inductor and capacitor, once current stops in the capacitor, the only option left is for the inductor will have to be taking that full 0.2 A as a simple conductor, as you mentioned, since current through it will no longer be changing. It will not be a "short" circuit because it is still behind that series resistor and so limited current.

With regard to your question as to whether the capacitor will discharge back through the inductor upon reaching maximum charge, the answer is no, and the reason for that is essentially the same as why a capacitor won't discharge back through a simple RC (no L) circuit: there's still a voltage drop, i.e. electric force, being applied from the voltage source and that will be holding it at maximum charge. For this case, remember that in a parallel connection, both components get the source voltage.

However, here's the interesting bit: if, on the other hand, you were to sever the wires connecting the inductor to the voltage source, then yes, indeed, the charge on the capacitor would now begin to flow through the inductor and it will discharge, as that holding force is no longer present. The inductor will then, as one may think, try to fight the resultant rising current, and it will do so on both the upswing and the downswing (as the capacitor approaches neutralization), meaning not only will it slow the capacitor's discharge, but it will at a certain point also try and "yank" charge back, and this will cause an oscillatory behavior: what you have now is a so-called "tank circuit". In theory, the charge will oscillate back and forth through the inductor forever, reversing the capacitor's polarity over and over again, however, in actuality, just as with many other conservative (i.e. not "creating energy from nothing" but rather simply keeping it in literal motion forever) "perpetual motion" schemes that seem to work in theory, the slight internal resistance in the wires due to being imperfect conductors will eventually dissipate all the energy into heat and the system will return to charge neutrality. Mathematically, the behavior of this circuit can be described as a damped harmonic oscillator, similar to a mass on a more realistic spring which is dissipating energy in internal friction due to bending the realistic material.

What you have is a filter. During the first part the cap will charge but unlike the inductor the cap is not an closed circuit Once it saturates there is no where for the emf to go. The coil provides a path for the current through the resistor so you will have current there. The interesting things happen when you apply AC power or open the switch. Then the EMF around the inductor collapses pushing current through the coil and the capacitor discharges, providing a temporary uninterupted power supply to the circuit. The effect is that, provided you reclose the switch fast enough, the circuit hardly recognizes that the switch was open.

This is very complex. At an instantaneous DC step change, current flows into the capacitor which causes the capacitor to charge as if it was charged through the resistor which it is. Voltage across the inductor causes the building of an electromagnetic field and initially counters the DC current, so initially, very little current flows through the inductor. As the capacitor charges and the inductor builds it's field, toward a peak, the capacitor and inductor combine to go a resonant circuit with the equivalent circuit of the capacitor, inductor and resistor all in parallel (assuming infinitesimal impedance at the battery). The circuit rings ( please don't say oscillates) and generates a damped wave at the resonant frequency determined by the inductor and capacitor with the number of cycles determined by the "Q" of the circuit derived from the inductance and the parallel resistance. Eventually as this cyclical ringing dies out, you have the steady state voltage across the capacitor and inductor equal to the voltage divider of the resistor and the resistance of the coil of the inductor (assumed to be infinitesimal). Current flows through the inductor based on the flow through the resistor and inductor resistance. However, opening the switch is a different story as the resistor is now taken totally out of the circuit and the quality factor of the circuit is increased and the number of cycles is limited by the resistances in the inductor and capacitor. Where closing the switch to produce a DC step change initiates a slightly reactive resonant circuit, opening the switch initiates a highly reactive resonant circuit which will ring with many more damped cycles than that with the circuit being charged. All of this can be determined exactly with the math but consider the charging and discharging circuits are very different with the elimination of the resistor.

If we assume that the variable voltage on the resistor acts to dampen any LC oscillations after the switch is closed (as indicated by the graph from MR. Centauri), then after a time, the voltage on both L and C will be zero and the L will have a current of .2 amps. If the switch is then opened, the resistor will be out of the circuit, and the energy in the inductor will oscillate between the inductor and the capacitor. You calculation for the maximum voltage under those conditions is correct.