# 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.