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Created | October 24, 2012 |

Last modified | October 24, 2012 |

Tags | energy imaginary-power inductor power power-dissipation power-loss real-power |

Currents, power and energy around a simple LR circuit.

Simulate > Time Domain > Run Time-Domain Simulation.

At t=0, I1 is pushing 1A through the parallel combination of L1 and R1.

L1 has zero resistance.

Because L1 has zero resistance, there is no voltage drop across it.

Because there is therefore no voltage across R1, the current through it must be zero.

Hence all the current flows through L1.

Therefore, at t=0, the total power dissipation in the circuit is zero.

From:

Eind = I^2*L/2

The energy stored in L1 is therefore 1Joule.

(http://en.wikipedia.org/wiki/Inductor)

At t=1s the current from I1 drops to zero.

I1 is an ideal current source so it has an infinite resistance.

The current in L1 cannot change instantaneously. Because it can no longer flow through I1, the current in L1 now has to flow through R1. This current flow through R1 now generates a voltage drop across R1. The combination of a current flow through - and a voltage drop across - R1 causes power to be dissipated in it. This converts the energy stored in L1 into heat and so removes energy from the circuit and transfers it into the surroundings by radiation and/or convection. Since there is a finite amount of energy stored in L1, converting some to heat means that the energy in the circuit reduces as time passes.

This is why the current in the circuit falls towards zero.

The time constant of this fall is L1*R1.

If R1 as well as L1 had zero resistance, the current would circulate indefinitely because there would be no power dissipation in the circuit.

This is what happens in superconducting energy storage rings:

http://en.wikipedia.org/wiki/Superconducting_magnetic_energy_storage

At t=10s, I1 switches back to sourcing 1A and the current in R1 steps up to 1A. The current in L1 cannot change instantaneously but climbs towards 1A with a time constant of L1*R1. As the current in L1 rises, the current in L1 falls until the circuit reaches the same state it was in at t=0.

In fact it is not exactly at the same point since this would take an infinite amount of time but for the purposes of this example this is a reasonable approximation.

The whole cycle then repeats over the next 20s.

This simulation also shows via V(integLpwr) and V(integRpwr) that the integral of the power apparently dissipated in L1 is in fact zero over a complete cycle of events whereas the integral of the power dissipated in R1 steadily increases with time.

This is because the energy removed from L1 in one half cycle is replaced in the next with no losses in L1 whereas the energy flowing through R1 results in the generation of heat which constitutes an energy loss due to a real power dissipation on every half cycle.

Note that due to the circuit not being allowed infinite settling time between I1 switching edges, there is asmall cumulative effect in V(integLpwr) and to a lesser degree in V(integRpwr).

(It is possible that some of the 'drift' in these integrals is also due to cumulative errors in the CL solver calculations.)

Note for the curious:

The expressions in I2 and I3 form 1 Ohm resistors which are used to zero the charges on C1 and C2 up until t=1ms. After that they operate as current sources generating currents equal to P(L1) and P(R1).

The powers have to be recalculated in these expressions because the P() plot expression function does not work inside a behavioural source expression.

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