Does a low resistance make it difficult to run a generator? Imagine I have a simple AC generator. I am providing energy to rotate the coil, which converts this to electrical energy. If I connect the ends to an electrical component such as motor, bulb, the power consumed would be equal to $P=V^2/R.$ Now if I instead connect the ends to a low-resistance component, the power consumed would increase, meaning that I would have to apply more power to rotate the coil at the same speed. What am I missing here?
 A: You are right. In the limit of infinite load resistance, no current is exiting the AC generator (the load is an open circuit) and it takes no work to turn the armature (except for overcoming bearing friction, etc.).
As the load resistance is reduced, progressively more current can then flow out of the generator and through the load- and more and more horsepower must be applied to the generator shaft to keep the armature turning at constant speed.
If you do not furnish more horsepower to the generator, the armature will slow down as the load resistance is decreased.
A: To expand just a bit on Niels' answer:
With less resistance in the load resistor, the higher the current in the coil. The higher the current in the coil, the higher the magnetic dipole moment of the coil. The higher the dipole moment of the coil, the greater the force between the coil and the magnet. The greater the force, the great the power necessary to maintain rotation of the coil against the force.
A: Imagine I have a simple AC generator which consists of a coil rotation in a magnetic field producing an emf $\mathcal E$ which is proportional to the angular speed of the coil, $\omega$.
Assume that the coil has a resistance $r$, the load resistance is $R$ and, to simplify matters, the frictional losses are small.
If there is a resistance $R$ across the terminals of the generator then a current $I$ flows in the circuit.
$\mathcal E = I(R+r) \Rightarrow \mathcal EI = I^2R + I^2r$
The $\mathcal EI$ term represents the mechanical power input to the generator, $I^2R$ is the useful power out and $I^2r$ is the wasted power in the coil of the generator.
Making $R$ smaller will mean that if the speed of the generator stays the same the current delivered by the generator will increase and so will the mechanical power input.
