The formula relating current to back emf is $V-E=IR$ where $V$ is the source pd and $E$ is the back Emf. Thus it can be seen that as the back emf increases the current decreases. But what is the implication of the power ($PV$) delivered across by the source and that across the compenent. I presume that both just decreases, otherwise energy would be lost, is this correct, and if so please can you explain?
Answer
Suppose we're dealing with a perfect motor, so there is no friction and no resistive losses. If we apply a voltage the motor will accelerate until the back EMF is equal to the voltage, and in this state we have:
$$ P = (V - E)I = 0 $$
The power is equal to the current times the voltage drop, but both the current and voltage drop are zero so no current flows and no power is dissipated. Now suppose we put some load on the motor, that is we get the motor to do some work. The load slows the motor, so the back EMF $E$ decreases, but the applied voltage remains constant so now:
$$ P = (V - E)I > 0 $$
and power is being consumed. Where is the power going? Well, it's going into the work done by the motor on the load we've applied to it. If the angular velocity of the motor is $\omega$, and the torque it's exerting is $T$ then the power is $\omega T$, so we have:
$$ (V - E)I = \omega T $$
The energy isn't lost, the motor converts it into work.
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