voltages and currents ???

Thread Starter


Joined Nov 4, 2007
hi, guys i m a bit confused with voltages and currents how are they related and affect any electrical device?

To start a device one needs some specific voltage say 220 volts
and current is equal to I=n*A*e*(vd)

where n is number of electrons(or density)
A is cross section of device through which it is flowing
e is charge on one electron
vd is drift velocity

now to operate a device one needs that current should flow through the windings of the motor or device

the internal impedance will affect the velocity of charge carriers (i m asking ur opinion here)

it will try to decrease the velocity of charge carriers

so, a kind of same voltage has to be applied across the conductor


if i see it logically ,the voltage cant remain the same through out as it appears to us , it has to overcome the internal impedance of the device so as to remain a constant drift velocity(m i right?)

how is power related to both of them ?
do devices really need constant drift velocities to run properly?

kindly help me out here people
need your help here to sort things out here in my brain?

thank you


Joined Jul 3, 2008
Good question. If I understand you correctly, the quick answer is that impedance is not really what changes drift velocity. The resistance is mostly a property of the material and device dimensions. The drift velocity is also depends on material, but it is proportional to both current and voltage. If we think of Ohms Law (V=IR), then the resistance can be constant if both voltage and current change proportionally. If we look at your formula for current: I=n*A*e*(vd), then Vd must change proportionally as current changes. So, voltage, current and drift velocity are all more-or-less proportional, and resistance is more-or-less constant. Also, power is P=VI which means it is proportional to Vd squared. -Kind of reminds me of kinetic energy. This is simplistic and approximate, but hopefully addresses the heart of your very good question.



Joined Jun 6, 2008

I am not sure whether the drift speed changes or not.

The voltage across a device powered by a voltage source (e.g. a battery if you wish), is what causes the charge to flow from a point of high potential (+ of the battery) to ground (- of the battery). In essence, voltage is merely a measure of how much potential energy each electron has (unit: joules/coulomb). Higher the voltage value, larger will the drift speed be. Lower the voltage, lesser the speed is. While the electrons pass through the device, they experience a voltage drop. This means, they lose some of their potential energy. Now where does that energy go? I have conflicting ideas as to where that energy goes, and there are two possibilities:

1. All of the energy (from previous paragraph) gets dissipated in form of heat, light, sound, or a mixture (whatever the device is), and the kinetic energy (therefore, the drift speed), remains unchanged. After a long time, the battery goes dead because its voltage (energy of each electron) has been continuously decreasing, and vd suddenly becomes zero (the device stops operating). In other words, vd is a discontinuous function of time right when voltage = 0.

2. Second possibility, some of the energy gets dissipated, while the kinetic energy of electron also decreases (meaning, vd decreases). So, vd is a continuous, decreasing function with respect to time.

Yes, the voltage of a source (e.g., a battery) must indeed be enough so that it supplies enough current to power the device (overcome its impedance), and not to forget, to overcome the internal resistance of the battery itself. Like Steve pointed out: Power = voltage * current.


I agree that the drift speed (vd) is proportional to current (I) based on the formula provided by Shankbond. However, I don't agree that vd is proportional to battery voltage (V).
U--> potential energy.
e--> charge of electron (1.6*10^(-19)C)
K--> kinetic energy (1/2)*me*(vdf^2-vdi^2); vdf--final drift speed,
vdi--inital drift speed

From law of conservation of energy,

U + K = dissipated energy (in the device -- heat, ligh, etc.)
V*q + (1/2)*me*(vdf^2-vdi^2) = dissipated energy

So, the voltage is related to the difference in squares of the initial and final drift speeds of an electron while it passes through the device.

This post is getting kind of too long. Can't help it, as Steve mentioned, this question really needs some in-depth research and analysis.

Thanks for sitting through the post.