We have a Low voltage Coming into our House as low as 80 Volts. Are we Paying more or less monthly?

Question

The Power Because of a Poor transformer can be 80 Volts
When we run the fridge for example or water Pump it all runs on Lower Voltage (Its Not Good) but is it costing us MORE or Less to operate these things or is it the same ? What is the relationship between Voltage and Kilowatt hours?

Answer 1

Jack says that you pay less because less voltage causes less current, then he gives the formula to prove himself wrong. Please don't give any more electrical advice.
With watts as a constant, a decreasing voltage causes an increase in amps and vice versa. You should be tripping breakers with much higher amps at 80V. I would be more worried about messing up your appliances that aren't made to run at that low voltage.

You are paying for watts per hour. It does not matter what your voltage or current is. Watts are relatively constant at any voltage. However, in resistive loads there is a nominal effect on power (watts) with respect to voltage.

Answer 2

You are charged by the kwhr. A so-called watt is a measure of energy expended. For example, a 60 W bulb uses twice as much energy as a 120 W bulb.

For DC, energy = W = I^2 R or = EI; where R is resistance in ohms, I is current in amperes, and E is voltage in, duh, volts. There are some differences in AC voltage, but they are similar; so we'll stick to DC, which is easier to explain the relationships.

Let's say your fridge is set to run most efficiently at 120 V. And it has a resistance R. Thus, it burns W = I^2 R watts when operating. I = E/R = 120/R. The reason it uses W energy is to keep things inside at a nice even temperature C = 0 degrees. (Wouldn't want your ice cream to melt in the freezer.) And that takes Wt = P power to do that; where t is the time the fridge is using up W energy.

From W = I^2 R = (E/R)^2 R = E^2/R, we find that P = Wt = (E^2/R)t is the power consumed by your fridge in time t, at voltage E, and with resistance R. Now we can look at what voltage does to power.

Take the ratio P/p = (E^2/R)T//(e^2/R)t/= (T/t)(E/e)^2; so that P = p[(T/t)(E/e)^2]; where E = 120 V and e = 80 V and P is power expended (kwhrs) at 120 V and p is power expended at 80 V.

Now set P = p; this results because it takes the same amount of power (kwhrs) to keep the fridge at 0 degrees C in the freezer. Then we have 1 = (T/t)(120/80)^2; so that t/T = (3/2)^2 = 9/4 = 2.25 or t = 2.25T. In other words, when you are getting e = 80 V instead of the E = 120 V you expect, the fridge needs to run 2.25 times longer than what it would normally run to keep your ice cream solid.

That relatively longer running time at reduced voltage will serve to wear out your compressor and compressor motor faster as the days pass by. In addition, because the motor would be running at a slower rotation rate, something called impedance in its coils could be considerably lower than designed. That in turn could allow more current than otherwise designed for. There is uncertainty on this last part because it depends on the motor design, which we are not privy to. But it could happen. And higher current means the coils could burn out.

To explicity answer your question, the relationship between voltage and kwhrs is P = Wt = (E^2/R)t where P is power in kilowatt hours, W is kilowatts of energy, t is time (hrs), E is voltage in you know what, R is resistance in ohms.

PS: Jack's answer is not entirely correct. You'd be charged less if and only if you could take the hit in power delivered. But if you need the same amount of power at 80 V that you needed at 120 V (e.g., to keep your ice cream frozen), then things would just run longer if they could run at all. Some appliances simply won't run at lower voltages, which is why some things stop during brown outs.

Answer 3

alternator is fine. check the two wires from the battery to the frame, the ground wire, and the starter, the hot wire. if they have a corrosion build up inside the wire they wont allow sufficient voltage to charge the battery at 13-14.7volts. next, take a voltmeter and check the dc voltage at the battery posts and then the post connectors with the car running and the positive post off the battery. car will keep running, but you will get a better voltage check of the alternator output when the car is running but the battery cable is off. if under 13 volts, replace the regulator if alternator is good, you may have a near-dead cell or bad terminal on the battery. if any corrosion is around a terminal replace the battery. also, check the water level in the battery itself. it may say maintenance free, but yoiu can still remove the caps and refill the water. low water prevents the battery keeping a charge and needs to be checked periodically. refill with distilled water, rather than tap water. the only reason they say a battery is maintenance free is to keep people from getting the full battery life because they go bad if they run out of water in the cells and the manufacturers don't want you to keep a battery as long as you are willing to buy another before you really need to.

Answer 4

motor or pump needs a certain minimum voltage to run, fine if you don't have such problem.

if voltage is lower, consumption wattage is lower, mathematically the relationship is as follows:

watt = ( V ^ 2 ) / Z

V is voltage
Z is the impedance or resistance of the load.

Answer 5

They have all missed the big picture, it will cost you a bundle, your appliances will have a shot life span, running on such low voltage.

Answer 6

It's costing you less. Less voltage causes less current.

Power (watts) = Current x Voltage

So they're both going down, which makes power go down by more.