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In this circuit, I calculate the current through R1 to be 10mA. What happens if I take into account the electromagnetic radiation in the ideal wire?

schematic

simulate this circuit – Schematic created using CircuitLab

edit: This is a pedagogical question, trying to establish that no, an ideal wire does not dissipate energy because that's not a useful definition of "ideal".

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    \$\begingroup\$ Then the wire wouldn't be ideal. \$\endgroup\$ – Dampmaskin Nov 1 '17 at 23:09
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    \$\begingroup\$ In this particular circuit, unless voltage or current is changing with time, there will be no electromagnetic radiation. \$\endgroup\$ – mkeith Nov 1 '17 at 23:30
  • \$\begingroup\$ @Dampmaskin: The title says it's just "an idea wire"! \$\endgroup\$ – Transistor Nov 2 '17 at 0:00
  • \$\begingroup\$ @Dampmaskin Good point. I've read that electromagnetic radiation in an ideal wire cannot be represented by inductors or resistors. Since this isn't an RF problem exactly (I'm interested in the DC steady state) I don't think a transmission line is the right symbol either. Any idea what the symbol for an electromagnetic-radiating, zero-inductance, zero-resistance ideal wire is? I know antennas have a characteristic impedance. So would it be an antenna with 0ohms impedance? \$\endgroup\$ – Cuadue Nov 2 '17 at 0:16
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    \$\begingroup\$ You have drawn 1V DC source. That has never been connected, it has been connected from t= minus forever and it will stay connected. So, there hasn't occurred and will never occur any state changes => No radiation, never! If there's some changes, you maybe edit them into the question. Ideal wires itself need some spec. If they have some length and thickness, they will have for example some capacitance. \$\endgroup\$ – user287001 Nov 2 '17 at 1:09
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If you accounted for the wires your circuit would look like the diagram below. It would have inductance from the magnetic field around the wire and resistance from the conductor.

The magnetic field effects are represented by the inductor and only apply if the current through the inductor is changing. Which in this circuit, once it is turned on the current is constant and the inductance does not matter.

If you had an ideal wire with electromagnetic effects, you could nix the resistors and just model the inductance.

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ There would also be some small amount of capacitance between (for example) the wires on the + and - sides of the circuit (exact amount unpredictable though--changes with the insulation around the wires, separation between the wires, and even (assuming there's an air-gap) the humidity and pressure of the air. \$\endgroup\$ – Jerry Coffin Nov 2 '17 at 3:17
  • \$\begingroup\$ And capacitance between the both wires and every other object in the room, but you have to stop somewhere. I also did not add the parasitics of the resistor. \$\endgroup\$ – laptop2d Nov 2 '17 at 6:07
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Well, it depends on the definition you give of "ideal wire" and the exact context you are involved into.

Usually in lumped element circuit theory (CT), a connection between two terminals is considered a piece of ideal conductor of zero length.

An ideal conductor is a conductor with zero resistivity, hence that connection can be considered like a zero ohm resistor.

In basic lumped element CT we assume the EM fields vary very slowly, so slowly to be almost stationary (i.e. constant with time). "Almost" here means that we can neglect all the terms of Maxwell's equations that involve time variations.

This also implies that any element in the circuit has physical dimensions that are much less than the wavelength of any signal component in the circuit. Moreover, it also implies that the whole circuit is much smaller than that wavelength (because we neglect the propagation delay of the signals in the circuit due to the finiteness of speed of light). In other words, in basic CT we assume that signals travel in the circuits with no delay other than those introduced by the lumped elements.

That assumption, together with the fact that those connections have zero length (well, to say things more precisely, they have dimensions much smaller than the lumped elements they connect), also imply that there are no parasitic effect.

All this is a really drastic simplification.

When you begin to relax your definition of "ideal", e.g. you assume connections are made with a conductor having some finite dimensions and non zero resistivity, you get what other have already said in this thread: parasitics.

In particular, some residual series resistance (due to non-zero resistivity of the material. Conductors having length also give you some stray inductance, because of them forming loops in your circuit and being coupled magnetically. Their proximity makes them develop stray capacitance between different parts of the circuit.

Moreover, if you begin to take into account also the dielectric that separates the wires (e.g. FR4 circuit board substrate) this both affects stray capacitance and introduces some parallel (leakage) resistance.

Another problem with "ideal wires" having non-zero physical dimensions is that the resistance of the conductor varies with frequency (no, I'm not talking about the equivalent impedance or its real part, just the raw resistance of the wire) because of the skin effect and the proximity effect (ouch!).

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  • \$\begingroup\$ +1 This is the correct answer. The question itself has nothing to do with radiation or RF, since everything is DC. \$\endgroup\$ – Shamtam Nov 3 '17 at 1:01
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All the radiation emitted by the power supply will travel parallel to the wires, then dive into the resistor, bending its path so it strikes the resistor surface at 90deg. With ideal wires, none is absorbed, but also none escapes.

For zero-ohm wires and any low frequency, the electrical energy travels like this:

poynting vector

A 2-plate battery with electrolyte on the left, is powering a grey resistive element on the right. The vector-field, showing the power-density and direction, is always perpendicular to both the e-field (shown as grey lines) and the b-field (not shown.)

Wires don't leak any radiation unless they're hundreds of miles long, or unless the power supply is RF rather than DC.

The model for radiation-leakage is the model for empty space: a vast array of capacitors and inductors chosen to produce c-velocity wave propagation. The array is connected at many points to your two wires. With pf and nH array elements, there won't be much leakage until you get far above 1MHz, or build an array hundreds of KM long.

Note that the e-field and b-field of this simple circuit actually looks more like the one below. Fill in all the planes, including distortions caused by the elbow-bends and the battery and resistor. Then sketch in the poynting-vector field perpendicular to those planes, so the energy-flow vectors connect all the junctions where the flux-lines of e cross the b. Also, more stuff.

enter image description here

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