# If the rotational speed of a turbine at a power plant doubles, does its voltage, current or power double? [closed]

Somehow, I cannot find a simple answer to this question, or a simple equation or equations...

All other things being equal, if a turbine at your local power plant instantaneously doubles its rpm (or, should I say, rps) what exactly happens to its voltage, current and power output?

I'm also curious about what happens with a regular DC Dynamo (brushed, commutator-bearing or brushless?) and homopolar generators when they double in velocity, too, but I'm primarily interested in a modern-day AC turbine....

Let's assume that turbine is constructed as the simplest dynamo machine and rotates with angular frequency $$\omega$$. Position of magnet in generator can be described the in terms of simple harmonic oscillator: $$x_{source}\sim sin\omega t\Rightarrow B_{dipole}\sim 1/(x_0-x)^2\approx B_0+kx$$ so $$U_{generated}\sim \dot{\Phi}\sim\dot{B}\sim\dot{x}$$ where for harmonic oscillator: $$\dot{x}_{max}\sim\omega x_{max}$$ So the voltage will linearly depend on angular velocity, the current will be different for different electrical circuit load. It doesn't mean you can get I=$$\infty$$ from the turbine, turbine will resist rotation with mechanical friction (low and doesn't depend on velocity) and magnetic resistance (the consequence of 3rd newton law, if you push electrons by field, they will push you), that will increase also linearly with $$\omega$$.
If dynamo machine works by rotating the coil of wire and not the magnet, you'll get the same result: $$S\sim S_0sin\omega t\Rightarrow U\sim\dot{\Phi}\sim B_0\dot{S}\sim\omega S_0cos\omega tB_0\Rightarrow U_{max} \sim \omega$$