At each commutation, energy is required to drive current into the phase that was previously unattached and is now attached to a power supply rail. The energy stored into the inductance is:

E=1/2LI^{2}.

Energy has units of Joules and a Joule equals a Watt*second. A Volt*Ampere equals a Watt.

The switched inductance is half the motor inductance so the energy stored in the inductor is:

E=1/4LI^{2}.

where L is the motor inductance.

Energy is required at every commutation, which occurs at a frequency of:

F_{com} = N[kRev/min]*1000/k*min/(60sec)*Poles/2[elec rev/mech rev]* 6/elec rev

F_{com} = N[kRev/min]* Poles*50

The power used to provide energy to the top switched inductance is:

P = E* F_{com }= 1/4LI^{2}*F_{com } = 1/4LI^{2}* N[kRev/min]* Poles*50

The average voltage used drive the switch inductance is.

V = P/I = 1/4LI* N[kRev/min]* Poles*50

This average voltage must be subtracted from the supply voltage in the equation for speed.

Substituting we get an equation with N on both sides:

N = (V-12.5*L*I*Poles*N-I*R)/K_{e}.

Solving for N:

N = 2*(V-I*R)/(2*K_{e}+25*L*I*Poles).

The average current supplied by the energy stored in the inductance released from the power supply rail is:

I_{ave} = P/V = 1/4LI^{2}* F_{com}/V

The average measured current is then:

I_{m} = (I-L*I^{2}*F_{com})/(4*V)

Next: Final comments on the motor model