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As you can see from the below graph, after an inductor is turned on, the current is stabilized to approximately 6.5mA. My question is: what does the 6.5mA represent? Where did that number come from?

enter image description here

When a capacitor is turned on, the voltage is stabilized to the source's voltage:

enter image description here

I can understand a scenario where the voltage of a capacitor and the voltage source do not match in voltage. But it doesn't make sense for an inductor and the source's current to not match in current. They have to match in current because they are in series.

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    \$\begingroup\$ What is your source current ? Is there a schematic to go with this ? \$\endgroup\$
    – efox29
    Sep 22, 2014 at 15:43
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    \$\begingroup\$ It stabilizes to the DC current of the load. In your case, the load was drawing 6.5mA. \$\endgroup\$
    – ACD
    Sep 22, 2014 at 15:44
  • \$\begingroup\$ @efox29: How about something like this circuit? i.imgur.com/K5QLx8a.png \$\endgroup\$
    – user53172
    Sep 22, 2014 at 15:58
  • \$\begingroup\$ Please correct the text of your question, it says '6.5A' in several places, but the current on the graph is '6.5mA'. \$\endgroup\$
    – gbulmer
    Sep 22, 2014 at 16:11

3 Answers 3

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Presumably, that graph corresponds to a specific circuit that eventually results in a 6.5mA current draw. But there is nothing special about 6.5mA. It's specific to whatever circuit that graph is describing.

You're expectation that an inductor in series with a source would have the same amount of current through it as the source is generating is absolutely true. I think your confusion stems from the fact that the inductor will actually affect the current coming out of the source.

For example, in Circuit A below, the current coming out of the source and through the resistor should be 6.5mA. Assuming ideal components, that will be true for all time. In Circuit B, however, the current coming out of the source will not be 6.5mA initially. The inductor actually "consumes" some of the energy in the circuit to form a magnetic field around itself, which manifests in slowing down the rise in current. And it will do so as long as the current is changing in time. Generating the magnetic field actually "resists" further changes in current. It's not 100% successful in resisting all of the current, it just slows down di/dt from rising instantly to 6.5mA. Eventually, however, the current out of the source and through the resistor and inductor will limit to the same as Circuit A, just as it does in the graph you posted. Once it reaches its maximum of 6.5mA, di/dt will equal 0. With no more change in current, the magnetic field will collapse and the inductor will act like a regular wire with no special properties.

schematic

simulate this circuit – Schematic created using CircuitLab

This result might seem to contradict Ohm's Law. Since the voltage of an ideal source can't change and the resistance doesn't change, how can the current change in time? Ohm's Law is satisfied because the voltage across the resistor changes over time. In any real circuit, there will always be resistance in the line, even if it's just the natural resistance of the conductor you're using. So the voltage source stays constant, but the voltage on the downstream side of the resistor adjusts to allow the current through the inductor to rise slowly.

Inductors have many uses, but one important use is as a filter. They inhibit sudden changes in current flow, which is helpful in filtering out transient spikes.

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  • \$\begingroup\$ Ah, this explains it perfectly. \$\endgroup\$
    – user53172
    Sep 22, 2014 at 16:13
  • \$\begingroup\$ I just edited and added an another paragraph (second to last) that I hope helps as well. \$\endgroup\$
    – Dan Laks
    Sep 22, 2014 at 16:17
  • \$\begingroup\$ "With no more change in current, the magnetic field will collapse and the inductor will act like a regular wire with no special properties." Does the magnetic field really collapse? Where does that energy go? \$\endgroup\$
    – user53172
    Sep 22, 2014 at 16:23
  • \$\begingroup\$ As the magnetic field collapses, the process actually reverses. The energy from the magnetic field actually induces current into the circuit. Inductors are very stubborn to change, no matter which direction. If di/dt is increasing, it absorbs some of the current into its magnetic field to resist the change. If di/dt starts decreasing, it releases energy from its magnetic field to encourage current to flow. This quality of inductors is the primary principle exploited in buck and boost converters. \$\endgroup\$
    – Dan Laks
    Sep 22, 2014 at 16:43
  • \$\begingroup\$ Ah, interesting. So if di/dt > 0, it absorbs current making the magnetic field larger and if di/dt < 0, it gives current making the magnetic field smaller and if di/dt=0, it holds the magnetic field. So I guess this was just a minor mistake? "With no more change in current, the magnetic field will collapse and the inductor will act like a regular wire with no special properties." \$\endgroup\$
    – user53172
    Sep 22, 2014 at 16:52
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At DC an ideal inductor behaves like a short circuit. The voltage across it is 0 but the current through it depends on the specific circuit it is in. In the case of your circuit the DC current is evidently 6.5mA.

An ideal capacitor has the opposite behavior -- it is an open circuit at DC. The current through it is 0 but the voltage across it depends on the specific circuit it is in.

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  • \$\begingroup\$ How about a circuit like this? i.imgur.com/K5QLx8a.png How would you get the current that it would stabilize to in this example? \$\endgroup\$
    – user53172
    Sep 22, 2014 at 15:59
  • \$\begingroup\$ In theory the current through the inductor would be infinite if the inductor and voltage source are both ideal since the voltage source is shorted. In reality the inductor has a small but non-zero resistance at DC and the voltage source cannot supply infinite current. You would need to know the inductor's non-zero DC resistance and the voltage source's maximum supply current. \$\endgroup\$
    – Null
    Sep 22, 2014 at 16:03
  • \$\begingroup\$ Alright, under ideal conditions, the current through the inductor would be infinite. But an inductor's current cannot change abruptly. So if it jumps from zero current to infinite current, then that is a problem. If it increases continuously from 0 to infinity and I=V/R and we know that I is increasing, then what is changing? V or R? \$\endgroup\$
    – user53172
    Sep 22, 2014 at 16:08
  • \$\begingroup\$ Assuming you have an ideal voltage source and ideal inductor the current will rise asymptotically to \$\infty\$ following the exponential curve as shown in the current plot in your question. But ideal voltage sources and inductors don't exist in reality so you would never actually see this in a real circuit. \$\endgroup\$
    – Null
    Sep 22, 2014 at 16:14
  • \$\begingroup\$ How about non-ideal conditions? The current of an inductor cannot change abruptly. So I goes from 0 to some current value. But as I increases, I=V/R so either V or R is changing. \$\endgroup\$
    – user53172
    Sep 22, 2014 at 16:16
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In this case, the 6.5Amps represents the current the resistive component of a REAL inductor to satisfy V = IR

It could also be the short circuit capability of the source.

Or a combination of both.

Shouldn't the question be 6.5mA and not 6.5Amps due to the y-axis scale?

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  • \$\begingroup\$ Yes, I believe "Shouldn't the question be 6.5mA and not 6.5Amps" too. \$\endgroup\$
    – gbulmer
    Sep 22, 2014 at 16:03
  • \$\begingroup\$ Sorry, 6.5mA not 6.5A. If we say that the I=V/R, then that causes a problem. Because we know that I is changing, V or R or both need to be changing. V can't change because it is a constant voltage source. R can't change, can it? i.imgur.com/K5QLx8a.png Here is a circuit example. I am assuming this is an ideal circuit and an ideal inductor. \$\endgroup\$
    – user53172
    Sep 22, 2014 at 16:04
  • \$\begingroup\$ Is that really the circuit? the circuit that was simulated? While R won't change it is constant. Say the Resistive component of the inductor is... 1500R (ish) then 6.5mA from a 10V supply would satisfy the circuit. Likewise source impedance. Basically more information because for an ideal inductor and an ideal power supply the current will rise to infinite. The plot you show has a time constant which implies some impedance somewhere \$\endgroup\$
    – user16222
    Sep 22, 2014 at 16:06
  • \$\begingroup\$ @user53172: V across the inductor changes. \$\endgroup\$ Sep 22, 2014 at 16:09
  • \$\begingroup\$ Sorry, there was no simulation. I am complete layman and the graph could be completely incorrect. From what I understood from the textbook though, the general pattern of the graph is correct when an inductor is hooked to DC. Ok, so if R is constant and V is constant, then how can the variable I change? \$\endgroup\$
    – user53172
    Sep 22, 2014 at 16:10

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