# If The Resistor Slows The Current By Reducing The Emf By 12v Then The

## If The Resistor Slows The Current By Reducing The Emf By 12v Then The

If the mechanical and electrical powers are correlated, as are the current and torque, then voltage and speed must be, also. And they are, because the faster the rotor spins through the stator field, the greater back-emf it will generate. This is Faraday's law of induction. So, …

Resistor Materials Resistors can be found in a variety of different materials, each one with its own properties and specific areas of use. Most electrical engineers use the types found below: Wirewound (WW) Resistors Wire Wound Resistors are manufactured by winding resistance wire around a non-conductive core in a spiral.

Let us assume above, that the capacitor, C is fully “discharged” and the switch (S) is fully open. These are the initial conditions of the circuit, then t = 0, i = 0 and q = 0.When the switch is closed the time begins at t = 0 and current begins to flow into the capacitor via the resistor.. Since the initial voltage across the capacitor is zero, ( Vc = 0 ) the capacitor appears to be a ...

It just slows it down. But the longer there's a resistor, it increases the probability that some of the electrons are going to bump into something and create a little bit of heat, et cetera, et cetera. So when you put resistors in series, what you're actually doing is increasing the probability that more electrons will bump into more things, right?

16/03/2018 · So if you have a slow switch (like a bipolar transistor) you will get a smaller back EMF compared to a fast switch like a FET (everything else being equal). A snubber is designed to provide a path for the inductance current to flow one the switch has opened, thus slowing down the change in current with time (di/dt), thus reducing the back EMF.

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18/10/2011 · Help with an electronic brake for a DC motor Home. Forums. Circuits and Projects ... so IR drop is 4.8V and back EMF is 7.2V. Then if you short the motor, you'd have 7.2/48 = .15A flowing, and that would be the peak. It would be less from then on. ... it gets closer to +12v. With the resistor size you're using for the NPN's base, you'd need to ...

One important point to note about the above equation. It only relates the emf produced across the inductor to changes in current because if the flow of inductor current is constant and not changing such as in a steady state DC current, then the induced emf voltage will be zero because the instantaneous rate of current change is zero, di/dt = 0.

19/03/2018 · Eventually it slows to the point where the current rises back to 10 amps and the original torque value is restored. As 60% duty cycle gives us about 28 volts, and the series drop at 10A is still 8 volts (as per original operation point), we can deduce that the motor will settle at whatever new speed gives only 20 volts back EMF.

Reducing the field strength is done by inserting resistance in series with a shunt field, or inserting resistances around a series-connected field winding, to reduce current in the field winding. When the field is weakened, the back-emf reduces, so a larger current flows through the armature winding and this increases the speed.