# Inductors and Capacitors

1. Nov 3, 2014

### needhelp171

1. The problem statement, all variables and given/known data
Explain what is happening in the following four circuits:

I don't know how to draw the circuits, but they're really really simple:

(a) Current source connected in a loop with an inductor
(b) Current source connected in a loop with a capacitor
(c) Voltage source connected in a loop with a capacitor
(d) Voltage source connected in a loop with an inductor

2. Relevant equations
i = C dv/dt (for capacitor)
v = L di/dt (for inductor)

3. The attempt at a solution

I feel like this is so easy but I have no confidence in my answers....

(a) The current source feeds constant current into the inductor. Note: CONSTANT current.

v = l di/dt
constant current ==> v = i * 0 = 0
Therefore there is no voltage drop across the inductor.

The problem is....I thought inductors and capacitors function as resistors in the steady state....which implies that there is a voltage drop across the inductor (since current is flowing through a resistor).

(b) The current source keeps pumping current into the capacitor. This causes electrons to build up on the plate. Thus, the capacitor is charging up. This can't go on forever because at some point the capacitor will fill up and not be able to take in any more electrons on its plate (after it's fully charged).

(c) The voltage source demands a voltage (v1) to go across it. This means there must be a voltage drop of v1 across the capacitor AT ALL TIMES. So, current runs through the capacitor until the capacitor gets charged to the point where the voltage across it is equal to v1. At this point current CEASES TO FLOW. And the capacitor is charged to v1 volts.

But, if that is the case, then what exactly is happening in the intervening time? That is, how is the voltage across the capacitor ALWAYS v1 if the voltage across the capacitor is constantly increasing until it reaches v1? Doesn't this imply that the voltage across the capacitor was BELOW v1 prior to that? Contradiction.....

ALTERNATE SOLUTION: The voltage is constant. Therefore, dv/dt is zero.
i = C dv/dt , where dv/dt = 0 ==> i = 0

The voltage source has a drop of v1 across a short circuit. Contradiction....

(d) The voltage is constant.
v(t) = L di/dt

To support the constant voltage, the current must be CONSTANTLY increasing forever and ever. This obviously can't go on forever, but that's what's happening in the circuit.

These are a lot more complicated than I thought...but I think if someone understands this they really have a solid understanding of inductors and capacitors. Please let me know where my logic has gone astray.

2. Nov 3, 2014

### BvU

If you don't worry too much about switching on, what you describe under a) is correct. And indeed, a real world inductor has some resistance.

Under b) the capacitor keeps filling (it can store Q = C V) so there's Always room for more charge, but the voltage across the thing increases as a consequence. Real world capacitors will break down with a discharge at some voltage.

Under c) you justifiably worry about what happens in time. Theoretically an infinite current charges the capacitor to V = Q/C in zero time. In reality there is some resistance to make it happen in finite time.

And d) is in the same ball park: if the inductor is ideal, the voltage source will explode, and if the voltage source is ideal, the inductor will have an unpleasant meltdown.

I think you get the picture pretty well. And your logic holds up. :)

3. Nov 4, 2014

### CWatters

What BvU said. Looks like you understand it all. Don't worry too much about real world components or potential problems such as "infinite" voltage or current. Best assume they are ideal components. In an exam if you have time you can expand your answer to include real world effects but chances are they just want the basics unless they say otherwise.