Kdjykuj

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 $0A$.

However, 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? Thanks!

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 $trightarrowinfty$), all the circuit voltages and currents are constant.

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

$$v_L = Lfrac{di_L}{dt}$$

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.

Tangential addendum:

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?

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

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:

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

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 is

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

Now when the switch is closed for a long time, the following rules apply to ideal capacitor and inductors.

The capacitor looks like an open circuit.
The inductor looks like a short circuit.

That means since the inductor looks like a short circuit, the current through the inductor is simply 10/50=0.2A.

Since the capacitor looks like an open circuit, the current through it must be zero. But the voltage will be 10 V.

In short, your book is correct.

Hope this helps.