Title: what is the voltage across the split ring commutator when the loop is rotating? Post by: lookang on November 03, 2009, 02:14:16 pm i am doing some research on java applet simulation on DC motor and AC generator.
http://www.walterfendt.de/ph11e/generator_e.htm show an interesting animation of the Dc voltage that resembled a full wave rectification. is that correct? i thought it will be a constant voltage. the reason is i am trying to create a Voltage vs time graph for http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=1266 and i am puzzled what shape it will be like? ohm law V =IR, during closed circuit, if I is constant, and R is a fixed quantity across the split ring, should V is constant? during open circuit, I = 0, imply V = V as it is the voltage of the battery so is it a constant line of voltage V = battery emf? or is the physics of electromagnetic induction at the loop more complex that and i forgot to consider? thanks for reply ;D Title: Re: what is the voltage across the split ring commutator when the loop is rotating? Post by: FuKwun Hwang on November 03, 2009, 08:58:23 pm I think the magnetic field is assumed to be uniform B=B_{0}.
The magnetic flux pass through the coil depends on the angle $\theta=\omega t$, assume the coil rotate at constant angular velocity $\omega$. i.e. $\Psi=\int \vec{B}\cdot d\vec{s}= B A cos(\theta)=B A cos(\omega t)$ From Faraday's law, induced voltage $V= \frac{d\Psi}{dt}= BA\omega \sin(\omega t)$ So the voltage is a sine wave (without commutator). Title: Re: what is the voltage across the split ring commutator when the loop is rotating? Post by: lookang on November 03, 2009, 09:33:34 pm I think the magnetic field is assumed to be uniform B=B_{0}. i agree with the above analysis that AC generator (without commutator) the emf induced in the coil is a sine wave.The magnetic flux pass through the coil depends on the angle $\theta=\omega t$, assume the coil rotate at constant angular velocity $\omega$. i.e. $\Psi=\int \vec{B}\cdot d\vec{s}= B A cos(\theta)=B A cos(\omega t)$ From Faraday's law, induced voltage $V= \frac{d\Psi}{dt}= BA\omega \sin(\omega t)$ So the voltage is a sine wave (without commutator). so is it that with a commutator, which flip the current in half cycle rotation, therefore emf induced in the coil will be a sine wave? if that is true, then i understand why the emf induced is sine wave. By the way, i am certain http://hyperphysics.phyastr.gsu.edu/HBASE/magnetic/motdct.html is accurate for torque vs time is a sine wave. (http://hyperphysics.phyastr.gsu.edu/HBASE/magnetic/imgmag/dcmtorv.gif) Title: Re: what is the voltage across the split ring commutator when the loop is rotating? Post by: FuKwun Hwang on November 03, 2009, 10:42:02 pm Yes. The commutator flip the current in half cycle to keep the voltage always positive.
For DC motor, there must be a commutator to generate torque in the same direction, otherwise, the ring will not rotate contineously (it will be become an oscillator.) Title: Re: what is the voltage across the split ring commutator when the loop is rotating? Post by: HY on March 05, 2010, 09:09:20 pm The adress of Walter Fendt's applet has changed. Now, it's http://www.walterfendt.de/ph14e/generator.htm On his page there are quite a few other very useful applets: http://www.walterfendt.de/ph14e/
