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#10558 06/15/02 08:03 AM
Joined: May 2002
Posts: 68
Eandrew Offline OP
Ohms law states "current is directly proportional to voltage". True, since more voltage will push more amps.

However, how am I getting this confused with the fact that everything in the real world suggests an inverse relationship between volts and current.

For example, a bigger load will decrease voltage. Or for high voltage transmission lines, the voltage is stepped up high so the lines can carry much less current and thus be very small.

where am I going wrong here?

#10559 06/15/02 08:53 AM
Joined: Aug 2001
Posts: 7,520

Yes, current is directly proportional to voltage, so long as the resistance remains constant.

Where you're going wrong is in comparing situations with vastly different values of resistance.

From Ohm's Law we have I = V / R.

We also have W = I x V.

Each of those formulae relates three of the four quantities involved, but you have to remember that all four are inter-related. That's how we can come up with other derived formulae, such as W = (I^2) x R.

Let's take an example:

A 60W light bulb on a supply of 120V will draw a current of:

I = W / V = 60 / 120 = 0.5A.

From that you can work out the resistance of the bulb's filament using Ohm's Law:

R = V / I = 120 / 0.5 = 240 ohms.

Now, if you were to increase the supply voltage a little, let's say to 140V, while that resistance remains constant both the current and the power would increase slightly:

I = V / R = 140 / 240 = 0.583A.

W = I x V = 0.583 x 140 = 81.7 watts.

(For pedants: Yes the resistance of the filament would change a little, but I'm trying to keep it simple!)

Similarly, if you lowered the voltage while the resistance stayed the same, both the current and the power would be correspondingly reduced.

Now let's look at a 60W light designed to run on a 240V supply, like we have here in England. Current will be:

I = W / V = 60 / 240 = 0.25A.

So here's where you see the proof that for the same amount of power (60W in this case), if you double the voltage you only need half as much current.

However, now work out the resistance of the filament:

R = V / I = 240 / 0.25 = 960 ohms.

That's a resistance four times greater than for the 60W 120V bulb.

So although we've doubled the voltage, the current is only half as much because the resistance is four times higher. The power has remained the same because 240V x 0.25A and 120V x 0.5A both come to 60W.


Your query about a bigger load reducing the voltage has a slightly different slant.

I think what you're talking about is the case where the measured voltage at an outlet drops when you connect or increase the load, yes?

What you're seeing here is not so much a reduction of the EMF, or source of the voltage, but rather the voltage drop in the wires to the outlet.

All wires have a certain amount of resistance, and obey Ohm's Law like any other component.

Let's assume that we start with exactly 120V and that the wires connecting to some outlet have a combined resistance of 0.5 ohm. If you draw a load of, let's say, 1A from that outlet, the the voltage dropped along the cable will be:

V = I x R = 1 x 0.5 = 0.5V

So the voltage measured at the outlet will drop to 119.5 volts.

Now if you increase the load to say 10A, the voltage drop along the cable becomes much higher:

V = I x R = 10 x 0.5 = 5V.

Now you're losing 5V in the cables so the voltage you measure at the outlet will be down to 115V. It's still 120V at the source, but you're losing 5 volts in the wiring.

The voltage drop on the wires is also why heavily loaded cables get warm. In this instance, the power dissipated by the cable would be:

W = I x V = 10 x 5 = 50W.

The trick with all of these calculations is to remember that Ohm's Law (and all the derived formulae) work with both the circuit as a whole, and with each individual part of the circuit. You just have to make sure that you use the right values for the part of the circuit in question, e.g. in the above power calculation, only 5V appears across the cables, so that's the voltage to use to work out the power lost in the cables.

The relationship between W, V, I and R is why you can use the other formulae to get the same result, e.g.

W = (I^2) x R = 10^2 x 0.5 = 50W.

Clear as mud?

Formula at top edited so that message reads properly. [Linked Image] Sorry about that!

[This message has been edited by pauluk (edited 06-15-2002).]

#10560 06/15/02 09:17 AM
Joined: Oct 2000
Posts: 4,116
Likes: 4

"We also have W = I x R." ??
... shouldn't that be I x V ? [Linked Image]

I would just like to add something for those that may be confused. I think that we are generally taught ohms law as I=E/R with E being Electro-Motive Force (Voltage) This is the same as the examples here with 'V' in place of the 'E' we may be more used to seeing. No difference (maybe ease of memorization?) When we sovle for Power (W in these examples) we use the formula P=I*E

We like our PIE here.

[Linked Image]

#10561 06/15/02 09:53 AM
Joined: Jun 2001
Posts: 29
Thanks Bill, thought I lost my mind for a minute, specially with the W(P)or(VA)=IxR formula. BTW I always considered E to equal Energy instead of EMF. ie: E=MC2. I've forgotten more than I should.

While you are at it Bill, is there a major difference between power (watts) and VA (volt amps)?

#10562 06/15/02 10:26 AM
Joined: Aug 2001
Posts: 7,520
Sorry about the slip-up. Darn fingers getting ahead of my brain again!

I wouldn't want to be accused of covering up a goof-up, but I've edited the original formula so that the explanation reads properly to avoid confusion.

The formulae here use W or P for power and V or E for voltage/EMF depending upon the source. Take your pick. Looks like I opted for the wrong ones to what most of you are used to seeing.

P = I x E only if the power factor is unity, but I didn't want to get into the complexities of reactive circuits.

Strictly speaking:

VA = I x E


Watts = VA x p.f.

Now, please pass the ice-cream to have with that PIE! [Linked Image]

[This message has been edited by pauluk (edited 06-15-2002).]

#10563 06/16/02 08:37 AM
Joined: Aug 2001
Posts: 7,520
<tongue firmly in cheek....>

Certain allegations have been made that I was involved in a relationship in which W = I x R.

I deny that I have ever been in such a relationship; this is just a vicious rumor put about by ^2, who was annoyed at being left out.

I have never been in or had any knowledge of any alleged relationship in which W = I x R. Except for that one time when I equalled 1, but that was pure coincidence and not my fault.

[Linked Image]

#10564 06/17/02 05:50 AM
Joined: Mar 2001
Posts: 118
Ohms laws looks great with DC but when you have AC and add some Xc and XL things turn into a bit of a Z :-)

#10565 06/20/02 05:10 AM
Joined: May 2002
Posts: 68
Eandrew Offline OP
The ohms law calculations were great. I printed them for future reference.

I think I was mixed up between:

transformer calculations and ohms law calculations.

With transformers, the inverse relationship exists. So if you step up the voltage, current goes down. and if you step down the voltage current goes up. This makes sence for transmission lines, (high voltage/low current) or big arc welder transformers (low voltage = high current.)

primary voltage/secondary voltage =

secondary current/primary current.

Maybe ohms law comes into play here too? Or maybe I'm ignoring the resistance in the number of turns of wire again?

I've studied this before but its a little fuzzy now

#10566 06/20/02 06:16 AM
Joined: Aug 2001
Posts: 7,520
Ohm's Law is such a fundamental concept that it comes into play virtually everywhere.

If you rearrange your Ep/Es = Is/Ip formula, you get:

Ep x Ip = Es x Is

As E x I gives you power (ignoring power factor), this is basically confirming that primary power = secondary power. In practice, of course, no transformer is 100% efficient, so secondary power will always be a little less than primary power.

I expect you're also familiar with the normal xfmr turns ratio formula:

Np / Ns = Ep / Es


To clarify the original problem, current is directly proportional to voltage, so long as resistance remains constant. Increase the voltage and you will increase both the current and the power.

The inverse relationship applies if we are assuming constant power, as in the 60W bulb example above. If we increase the voltage, we need to decrease the current to keep the power the same. The only way to increase E and reduce I simultaneously is to increase the resistance.

[This message has been edited by pauluk (edited 06-20-2002).]

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