# Electronic – Energy conservation when adding braking resistor to H bridge

dcenergyh-bridgepower

This may be a stupid question but I'm missing something fundamental about energy conservation of electrical machines – I am sure I have forgotten something from basic machine theory but I haven't found out what thus far:

Let L, R, E be the equivalent model of a DC brushed motor winding (inductance, resistor, back-EMF).

Say we are braking the motor either by short-circuiting the windings, or applying the reverse voltage, let I be the current value. The mechanical power is -EI right? Energy conservation dictates this power is dissipated somewhere, and here it's RI^2.

Now, I close a switch and insert in series with the winding an additional braking resistor (for example when freewheeling the current goes back to the supply and into a braking resistor when the DC link regulator switch is closed). The braking power is still the same, -EI, since the current has not yet changed thanks to the inductor. However, now both the windings and the braking resistor dissipate more power than before the switch closed: (R+R2)I^2.

What am I missing? Is the inductor providing the power for the additional resistor – meaning the energy dissipated in the winding cannot be reduced for a given mechanical energy to be dissipated?

This is ultimately what I am trying to achieve, my winding cannot handle the mechanical energy I have to dissipate from the rotor when braking.

## This is for inductor stored energy

So, if you discharge a charged inductor onto a 1 Ω resistor, the instantaneous current will be the same as you connecting the inductor to a 1000 Ω resistor but, the whole show will be over much more quickly. In other words, with a 1 Ω resistor, it might take 1 second for the current to reach a level of one-tenth whereas for the 1000 Ω resistor it will take much, much less time and...

...the total energy taken by both resistors over a long time period will be the same. They will both get warmer by the same amount (assuming perfect heat insulation).

The energy that either resistor receives is $$\\dfrac{1}{2}\cdot LI^2\$$ where $$\I\$$ is the instantaneous current at the point when the resistor is connected to the inductor and, that current is defined only by the inductor (as we know).

## For energy due to mechanical momentum

This is different and there is no correlation as implied here: -

The braking power is still the same, -EI, since the current has not yet changed thanks to the inductor.

So, don't mix up the two.