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If an inductor is ideal then it has no resistance, but only an inductance, how does the short-circuit affect the inductor? Why would the voltage across it be zero?

I know that the 6H by mutual induction would cause a current to go into the 10-H inductor but how would it work if we assumed for just a bit that the 10H is alone

  • \$\begingroup\$ However, it does have reactance. \$\endgroup\$ – Gunnish Nov 20 '13 at 8:42

While an "ideal" superconducting inductor would have zero resistance, it does have an impedance, which is a function of the frequency of the driving signal, and thus the current flowing across it.

Put simply, "An inductor opposes any change in current through it".

The circuit needs to be examined in an AC signal model, in which the inductor is not a short circuit at all.

Thus, while a DC signal across the hypothetical ideal 10H inductor would see a short circuit, the supply shown in the diagram is an AC signal of 24 cos 4t Volts.

By computing the impedance of a 10 Henry inductor at the specified frequency, the actual current through it can be determined, and this will not be infinite - it will decrease with increasing frequency.

Edit: Missed the straight wire across the 10H coil.

In the situation where there is a zero-ohm, zero inductance straight wire shunting across the 10H coil, the concept becomes simpler to understand:

The reactance of the wire is zero, the resistance of the wire is also zero (ideal), and since V = I x R, therefore the voltage across that piece of wire = 0 for any defined current io flowing through it.

Since the voltage between those two points is zero, that's also what is across the inductor, or any other component spanning those two points.

  • \$\begingroup\$ If the 6-H inductor was not present in the circuit, would all the current flow into the short? If we assume that it there, how is the voltage across the vertical coil zero? \$\endgroup\$ – user29568 Nov 20 '13 at 8:43
  • \$\begingroup\$ My answer takes into account that the 6H inductor is not present: The 10H inductor is still an inductor, so it has reactance that varies with frequency. Voltage across the inductor is not zero, it will be the same voltage as the AC supply shown. What makes you believe it is zero? \$\endgroup\$ – Anindo Ghosh Nov 20 '13 at 8:47
  • \$\begingroup\$ My professor was explaining it on the board and he said the the voltage across the vertical coil is zero.. \$\endgroup\$ – user29568 Nov 20 '13 at 8:48
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    \$\begingroup\$ Ahhh hah, I misread the schematic, I missed the wire across the ends of the 10H coil. Well, if there's that straight wire there, then no matter what supply is provided, the voltage across that straight wire will always be zero, since V = I x X (reactance), and X for an ideal straight wire = 0 \$\endgroup\$ – Anindo Ghosh Nov 20 '13 at 8:50
  • \$\begingroup\$ And since it is in parralel with the 10-H inductor the voltage across it is also zero \$\endgroup\$ – user29568 Nov 20 '13 at 8:52

A short circuit doesn't affect the inductor. It remains an inductor and continues to operate within its mathematical boundaries. If there were current flowing through the inductor when the short was applied this current will continue for ever but diminish towards zero amps if there are losses such as non-ideal zero ohms resistance.

It's different with a capacitor; if there were a voltage across it and you applied a short circuit, infinite current would flow.

An inductor is similar to a capacitor when current is flowing and it is open-circuited - it theoretically will produce infinite voltage.

  • \$\begingroup\$ So basically the short-circuit is superfluous since the current that enters the short circuit is equal to the current that enters the 10-H. \$\endgroup\$ – user29568 Nov 20 '13 at 8:36
  • \$\begingroup\$ No, the current through the short circuit due to the voltage source in series with the 6 H inductor will obviously flow through the short but also the current that was flowing through the 10 H inductor will add to it. \$\endgroup\$ – Andy aka Nov 20 '13 at 8:38
  • \$\begingroup\$ I got lost. If we assumed that the 6-H inductor is not in the circuit, how would the current flow from the voltage source \$\endgroup\$ – user29568 Nov 20 '13 at 8:40
  • \$\begingroup\$ without the 6H inductor there would be no current in the 10H inductor and therefore no current in the short circuit. Nothing, it would be a beyond-trivial circuit that does absolutely nothing. \$\endgroup\$ – Andy aka Nov 20 '13 at 8:43
  • \$\begingroup\$ How can there be no current? And if we assume that it is there, how is the voltage across the vertical coil zero? \$\endgroup\$ – user29568 Nov 20 '13 at 8:46

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