Motor Efficiency: Increase in Voltage & Why?

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The discussion focuses on the efficiency of electric motors and how it relates to voltage levels. It highlights that efficiency can increase with higher voltage, but this is a misconception that needs clarification. Factors affecting motor efficiency include various loss mechanisms that prevent 100% energy conversion. The conversation emphasizes the importance of understanding different motor types and their specific characteristics. Addressing these points is crucial for accurately completing the A-level coursework.
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for my A level coursework i am doing the efficieny of a motor, I've just about sussed the theory:confused: . Now i have found out that the efficieny increases as the voltage increases and its wrong! :mad: i need help on saying why this would be the case. help would be much appreciated!
 
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Welcome to the PF, heena1988. Homework and coursework questions need to be posted here in the Homework Help forums (where I've moved your thread). And you need to show us some of your own work and thoughts in order for us to help you.

So, what factors go into the efficiency of an electric motor? There are many different types of electric motor -- which type(s) are you asked about? What are the loss mechanisms in motors (there are several) that keep the motor from being 100% efficient in converting electrical energy to mechanical energy?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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