Voltage Drop Under Load

Okay. Thanks to everybody who answered my question. I have a much clearer picture of what is going on. I'm going to start taking a space heater around with me so I can measure the length of the runs to the outlet I'm at just for the fun of it :)

-- Jack

Do you know the wire size? Also you need to know the voltage at the circuit breaker panel with the heater plugged into the outlet and then at the outlet to calculate the wire length.

Right.

Baloney. Your concept and your math are wrong.

All he needs to do is measure the voltage at the receptacle without the heater, and that is the voltage at the panel. Then plug in the heater and note the value under load. The difference is the voltage drop.

Ben Miller

Thats rather clear...thanks... the voltage drop is a function of the static reistance and the amps.

Phil Scott

------------ Ben Miller has it right. Resistance is a property of the material.

Please note that, for a pure ohmic resistance, ignoring thermal effects we have R=V/I. You are saying that for I= 0 then R= 0. However for a non-zero voltage V the relationship implies that R must be infinite if I is 0. This is contradictory to your statement. If the voltage is also 0 then the statement becomes R=0/0 which is NOT necessarily 0 and could be any value (and of no interest) as 0/0 is undefined.

Now you give an example of a motor which measures 3 ohms with a hand held ohmmeter - which does NOT have a 9VDC current-but a 9VDC voltage source behind an internal resistance- the current is a few ma. The actual current drawn in the case of a motor is not determined by the resistance of 3 ohms but by the mechanical load on the motor as reflected to the electrical side- the motor presents a generated or back emf in series with the winding resistance- it is an active load. The ohmmeter result does not allow determination of the current by Ohm's Law because, with an active load, such as a motor, Ohm's Law is useless garbage. Ohm's Law, strictly speaking, is true for a motor only when the shaft is locked so it cannot turn. In such a situation, the motor is useless. I suggest that you review Ohm's Law definition and applicability.

Your last paragraph has poor math as 0/0 is not defined. It could be 0 or could be infinite or anywhere in between. Secondly, the last statement in this paragraph appears somewhat garbled. Any current, whether above or below the rating of the conductor will cause a voltage drop between source and load. The larger the current, the higher the drop - not because the resistance increases but because there is a V=IR drop in the conductor even for R constant- leading us back to the original statements given correctly by Ben and others.

Yes, that will work, good call. ( although, if you know the voltage is equal at both ends, then you know nothing else is drawing current from that run.) Mike

Ohm's Law V=IR. Assume the resistance of the wire is 0.1 ohm per 100 ft. If you have 100' of wire with a 1 amp load, the voltage drop over the length of the wire is 0.1 volts. If you have 120 volts at the panel, you would see 119.9 volts at the other end of the wire. If your load is 10 amps, you will have a 1 volt drop and see 119.0 volts. If you had 200' of wire with a 10 amp load, your drop would be 2 volts, and you would only get 118 volts.

In addition, if you really overload the wire it will in essence become a heating element. The resistance will increase, increasing the incremental affect of additional load. It's even worse with a motor as the load, since an insufficient voltage supply to the motor will cause it to consume excessive current. A common problem is ot see a motor trip a breaker when on a long length of small wire, such as a drop cord.

Good analogy. For some reason, everyone understands plumbing, but no one can grasp electricity. I guess it is because you can see the water. So are we in agreement that a water balloon is like a capacitor, and a siphon is like an inductor? So what would be the electrical circuit for a toilet?

Good Question. I think I would have to revert to Direct Current. The tank being a big capacitor. The valve being a Transistor or better an SCR (Silicon Controlled Rectifier). I guess the line coming in would amount to some sort of resistance. So fire the SCR, it dumps the tank. When the tank is empty the SCR shuts off, and the tank slowly fills again. That is just about what happens with a cars capacitive discharge ignition.

I think you need to go back to school. You are talking nonsense lad. I doubt you can even change a light bulb.

sQuick..

It's not quite that easy! You need to know the resistance of the space heater, since it will not draw the same current at which it was rated, with any voltage drop. It's resistance is its rated voltage squared divided by its rated wattage. This will be a constant, (in spite of what Phil Scott says), neglecting the slight difference in temperature at the lower voltage. Then, the resistance of the wiring to the outlet is simply the difference in the loaded voltage and the unloaded voltage, divided by the loaded voltage, all times the heater resistance.

In the USA, almost all outlets are wired with #12, (20A), or #14, (15A) wire. Copper wire of these sizes has a resistance per thousand feet of

I wouldn't say a siphon. When explaining it to folks using the 'plumbing analogy', I've likened an inductor to a water-wheel in the pipe with a very massive rotor. Once the water gets the wheel spinning, the inertia of the wheel will actually try to 'push' the water through the pipe after shutting off the supply.

A resistor (supply pipe) from the line charging a capacitor (storage tank)[the float valve needn't be considered, but could be]. On the downstream side of the capacitor, a pushbutton with some low resistance in series (flapper valve) supplying another low resistance in parallel with a zenor diode. Of course, if the drain line (low resistance) gets 'plugged up' (higher resistance), then the zenor conducts water all over the floor :-)

Now, imagine a spring-loaded check valve that has a small latch that will keep the valve closed until a solenoid 'unlatches' it. Then the valve stays open as long as there is forward flow through it. If the flow drops too low, or tries to reverse, the check valve slams shut and re-latches again. What's the electronic equivalent?? :-)

daestrom

By your statements then how do you apply Ohm's law? If the current through a wire doubles, how does the resistance change? You say it is zero at zero current, what does the resistance do if the current is changed from one ampere to two ampere? It remains exactly the same, it does not 'increase with current flow'. Only the voltage drop through the wire increases.

Not it is *not*. If E=IR and you set I and E equal to zero, then R can have any value in the universe (even negative) other than infinity and the algebraic equality is still satisfied. And for most materials, the value of R is a constant for the material if the temperature and physical geometry of the material is held constant. Regardless of the magnitude of I (neglecting skin effects and extremes that melt/vaporize the material).

daestrom

An Silicon Controlled Rectifier or a Thyratron. Once triggered on it continues to flow until the current is about zero.

An SCR!

Ben Miller

BINGO!!! We have a WINNER!!! :-)

daestrom

Only eight minutes after the first entry, another WINNER!!! ;-)

daestrom

Or just water flowing in a pipe. If a valve at the end of the pipe is suddenly closed it produces a pressure (voltage) spike and water hammer.

bud--