I know that a capacitor stores charge:
C=Q/V
but what i don't understand is how this would reduce the voltage drop caused by high current draw.
My theory is that the capacitor would need to be in parallel with a conductor (minimal resistance) and then connected to the motor.
simulate this circuit – Schematic created using CircuitLab
(Diode is motor)
By doing this, current will only be drawn from the capacitor when the power source can't supply enough.
If this is true why? and if it this isn't true how is capacitor supposed to be set up then? what would happen if the capacitor is in series?
Best Answer
What you show is just a diode in parallel with a voltage source. There is no capacitor here since it's shorted. Removing the capacitor would change nothing.
It's not clear what you are really asking, but some types of motors have a "startup capacitor" in them. These types of motors run on AC, and don't have any torque when the rotation speed is 0. The capacitor unbalances the motor to cause some torque at 0 speed.
This capacitor would reduce efficiency at normal operating speed, so there is usually a way to switch it out of the circuit. A common means is a centrifugal switch.
Added
If you really mean a capacitor in parallel with the power supply a DC motor is connected to, then that's just a capacitor holding up a supply. There is nothing special about a motor being connected to that supply.
A large capacitor across a supply provides extra charge to the load when the supply voltage drops. This helps the supply look more beefy to the load than it really is, at least in the short term. In effect, the supply/capacitor combination is capable of larger short term current than just the supply alone.
A motor draws a surge of current at startup, so a capacitor can help. However, this motor initial current surge is "long", so in most cases a unrealistically large capacitor would be needed to make a significant difference.
A small capacitor across a motor can help to reduce emissions. The capacitor keeps the voltage more steady, and keeps the high frequency noise current circulating close to the motor. The time over which such a capacitor can make a meaningful difference in holding up the voltage is so small that this only does anything useful at frequencies that can radiate.