Power equation in railgun operation.

[avicenna]

In a capacitor discharge through a plain resistor, the capacitor power supplied at any instant is VI; the power dissipated in the resistor is I²R. So VI = I²R.

Consider a railgun operated with a capacitor bank. At any instant of capacitor discharge, the power supplied is VI. The total power supplied for ohmic loss is sum I²R for two rails plus the resistance of the armature.

Question: Since VI = total I²R, how can the power equation include the kinetic energy supplied to the armature?

 

[exchemist]

33 minutes ago, avicenna said:

In a capacitor discharge through a plain resistor, the capacitor power supplied at any instant is VI; the power dissipated in the resistor is I²R. So VI = I²R.

Consider a railgun operated with a capacitor bank. At any instant of capacitor discharge, the power supplied is VI. The total power supplied for ohmic loss is sum I²R for two rails plus the resistance of the armature.

Question: Since VI = total I²R, how can the power equation include the kinetic energy supplied to the armature?

 

I should imagine it will be the same as any electric motor. In the stall condition all the power goes into ohmic loss, but as the motor picks up speed a back e.m.f. develops and it becomes more complicated. So I think the answer is that VI = I²R no longer describes the situation. But I'm very rusty on this: the last time I really studied it was for A Level in 1971.  I seem to recall that the power of the motor is actually given by the back e.m. f multiplied by the current, so in total you end up with: VI = EI + I²R. 

But no doubt someone will correct me if this is wrong.

Edited April 14, 2022 by exchemist