ohms Law question

[POWER_PIG]

Im sitting here staring at my study guides for the Masters Exam and Im confused how it can be stated that voltage and current are directly in proportion to one another (increase voltage by 25% and the current is also increased 25%). I can flip a few pages and it states that an increase in votage will decrease the current. I know Im missing something and it has to do with fixed resistance, but im having a hard time understanding. Any help will be appreciated. :-?

 

[bphgravity]

1 volt will push 1 ampere through 1 ohm of resistance...

2 volts will push 2 amperes through 1 ohm of resistance...

10 volts will push 10 amperes through 1 ohm of resistance...

This is a directly propotional relationship between voltage and current.

10 volts will push 20 amperes through .5 ohms of resistance...

10 volts will push 5 amperes through 2 ohms of resistance...

This shows the inversely proportional relationship between current and resistance.


Now consider an occupancy with 50,000va of load requirements.

A 50,000va load at 240V, 1? will require a capacity of just over 200A.

A 50,000va load at 480V, 3? will require a capacity of just over 60A.

To provide the same power requirements, increasing the voltage decreases the amapcity of the system.

 

[dereckbc]

Either your text book is wrong or you are missing something. If the power remains constant say a 400-watt light fixture, then by choosing a higher operationg voltage, say 120 to 277, the current will be lower. However if the resistance remains constant, then any increase in voltage will increase current and power.

Just do the math and you will see.

 

[infinity]

Im sitting here staring at my study guides for the Masters Exam and Im confused how it can be stated that voltage and current are directly in proportion to one another (increase voltage by 25% and the current is also increased 25%). I can flip a few pages and it states that an increase in votage will decrease the current. I know Im missing something and it has to do with fixed resistance, but im having a hard time understanding. Any help will be appreciated. :-?

In a resistive load the current and voltage are directly proportional. For an inductive load they are inversely proportional. A dual voltage motor will have a current of about 1/2 when the voltage is doubled.

 

[hillbilly]

I'll try to make it simple.

Say you have (2) 100 watt light bulbs, each rated for different voltage... bulbs that produces 100 watts of heat (a watt is a unit of heat).

If the bulb is rated for (and operating at) 100 volts, and it is producing 100 watts power, the curent flow thru the circuit will be 1 amp. The circuit resistance will be 100 Ohms.

If the bulb is rated for (and operating at) 200 volts and it is producing 100 watts power, the curent flow thru the circuit will be .5 amps. The circuit resistance will be 400 Ohms.

As long as the...POWER (watts)....REMAINS THE SAME,....... any.... INCREASE IN VOLTAGE.....will.... RESULT... in a proportional....DECREASE...in...CURRENT... flow.

or

If you have a circuit that is running thru and powering a ...FIXED RESISTANCE... (oven element for example, could be a light bulb), any INCREASE in VOLTAGE.... will result in a
proportional ...INCREASE...in....POWER...

Say you have a 10 Ohm resistor....
If you apply 100 volts, that resistor will produce 100 watts of heat and the current flow will be 1 amp.
If you apply 200 volts, that same resistor will produce 200 watts of heat and the current flow will still be 1 amp.

In both examples, the resistors in the circuits are fixed and in both examples, and The current (amps) and Power (watts) is dependant on Voltage.

I think that your confusion may be in the understanding of...POWER... verses... CURRENT... and you can't determine POWER without knowing the voltage of the circuit.

Just because you have large or small CURRENT flow (amperage) doesn't mean that you have large or small POWER output....it's DEPENDANT on VOLTAGE.

These are simple circuits and (using Ohm's Law) you can calculate Current, Voltage or Resistance if you know the other two.
To calculate Power, you must know all three.

Hope that helps.
If I've made any mistakes, I'm sure that we'll both find out.
steve

 

[rattus]

Power/Energy Again:

Power/Energy Again:

I'll try to make it simple.

Say you have (2) 100 watt light bulbs, each rated for different voltage... bulbs that produces 100 watts of heat (a watt is a unit of heat).


If I've made any mistakes, I'm sure that we'll both find out.

Hillbilly, you made a mistake.

A watt is a unit of power.

A watt-hour is a unit of energy.

Heat is measured in units of energy, i.e., BTUs or calories.

One KWH = 3413 BTUs

If you burn the 100W lamp for 10 hours, you transform 1 KWH of electrical energy into 3413 BTUs of thermal energy (heat).

 

[Smart $]

If I've made any mistakes, I'm sure that we'll both find out.

That's a given in this(these) forum(s).

Aside from what rattus brings to the table, I see a couple others in your post.

First:

If you have a circuit that is running thru and powering a ...FIXED RESISTANCE... (oven element for example, could be a light bulb), any INCREASE in VOLTAGE.... will result in a
proportional ...INCREASE...in....POWER...

The use of the word "proportional" here is somewhat ambiguous in that the increase in power is exponentially proportional and not directly proportional.

P = E?/R​

Second:

Say you have a 10 Ohm resistor....
If you apply 100 volts, that resistor will produce 100 watts of heat and the current flow will be 1 amp.
If you apply 200 volts, that same resistor will produce 200 watts of heat and the current flow will still be 1 amp.

Ummm... if you apply 100 or 200 volts across a a 10 ohm resistor the current flow will be 10 or 20 amps, respectively... not 1 amp... and the power consumed will be 1000 and 4000 watts, respectively.

 

[26AJ]

Say you have a 10 Ohm resistor....
If you apply 100 volts, that resistor will produce 100 watts of heat and the current flow will be 1 amp.
If you apply 200 volts, that same resistor will produce 200 watts of heat and the current flow will still be 1 amp.

Don't you mean that 100 volts applied to a 10 Ohm resistor will produce 1000 watts and push 10 amps?

E=IxR

 

[LarryFine]

Say you have a 10 Ohm resistor....
If you apply 100 volts, that resistor will produce 100 watts of heat and the current flow will be 1 amp.
If you apply 200 volts, that same resistor will produce 200 watts of heat and the current flow will still be 1 amp.

If I've made any mistakes, I'm sure that we'll both find out.

Yep, above. With a constant resistance, doubling the voltage will double the current, and result in a quadrupling of the power.

Ohm's Law states that one volt will push one amp through one ohm.

E = I x R
I = E / R
R = E / I

Watt's Law (name debatable) states that one volt at one amp produces one watt of power.

P = E x I
E = P / I
I = P / E

The rules: If the resistance is constant, the current varies proportionately with the applied voltage. If the current is constant, the resistance varies proportionately with the voltage. If the voltage is constant, the current varies inversely with the resistance. Current is the rarest number to be a constant.


Here's what is confusing the OP: when you want to juggle Ohm's Law figures, you first have to determine what will remain constant in your example circuit. There are the constants (the knowns) and the variables (the unknowns) In most real-world circumstances, it is the resistance (the equipment) and the voltage (the supply) that are known.

You really have to determine what are the constants in your given situation; you must know at least two parameters. As I said, the system voltage and equipment impedance (deduced from design voltage and rated current) are the known's. If you give us a couple of examples, we can show you using your numbers.

 

[hillbilly]

POWER_PIG....I really opened the flood gates and got you some answers didn't I?

steve

 

[LarryFine]

If you burn the 100W lamp for 10 hours, you transform 1 KWH of electrical energy into 3413 BTUs of thermal energy (heat).

Doesn't that leave out the portion of energy that went into creating visible light?

 

[rattus]

Well, yes!

Well, yes!

Doesn't that leave out the portion of energy that went into creating visible light?

I expected that response from someone. What if I said radiant energy?

PS: Larry, I think the visible energy is absorbed and converted to heat anyway.

 

[POWER_PIG]

Thx All

Thx All

You sure did Hillbilly!!! LOL!
Many thanks to all, I appreciated the swift responses.
All the info has been a huge help!!

 

[POWER_PIG]

still bummin

still bummin

what is confusing me is the fact that a typical home is full of resistive loads, yet the utility company greatly reduces current by raising the voltage on the grid supplying the load.
On the other hand if I had a load with a fixed resistance of say 3.2 ohms, the more voltage applied will result in a higher current
I spose the part that is still bothering me is that how can a resisitive load be fixed?
Say its Thanksgiving morning and 50 houses on one single phase 7620v primary distrubution grid have their 10kw 240v ranges running at full capacity at one time. (assuming only the ranges on and 100% eff on the transformers for simplicity) well, 500,000/7620 = 66amps apx.
Well now lets take just one house and apply say 480 volts to a range rated 10kw 240v. First I'll find the fixed R. 240X240/10000=5.76 ohms and 43a.
now lets slap some 480v on that same range. 480/5.76=83a. Wow! what if that same 7620v came ripping through those elements.......7620/5.76=1323a
thats where I get stuck......:roll: thx again for all you help

 

[tallgirl]

You're mixing some concepts in here.

The POCO reduces the current by increasing the voltage -- that much is true. However, that's just so that P is a constant, while VD is reduced.

That is, P = EI and VD = IR, so by increasing their voltage through step-up transformers, the POCO reduces the drop in the voltage over the distribution network. We've done this discussion before, but briefly, if you double E, you halve I. Since you've halved I, you halve the numerical value of VD, which results in 1/4th the percentage. Here's a worked example --

E = 120v, I = 100A, R = 0.1 ohms. At 100A a 0.1 ohm load results in a 10 volt drop. So, VD = 10v, and the voltage on the other end of the wire is 110v.
E = 240v, I = 50A, R = 0.1 ohms. At 50A a 0.1 ohm load results in a 5 volt drop. So, VD = 5, and the voltage on the other end of the wire is now 235v.

As you can see, the percentage voltage drop has been greatly reduced.

Now, you get to the question of loads and how E and I are related. When you have a fixed value of R, which you do for purely resistive, non-reactive loads, I = E/R. Increase E and I increases in direct proportion. That's Ohm's law. If you increase R, I decreases in inverse proportion. Again, Ohm's law.

Remember -- the POCO used a transformer to go from (say) 120v @ 100A up to 240v @ 50A. In reality they are going to use much higher voltages that are harder for me to do simple math with

When you get to an electric oven, the heating element is a purely resistive load. So, 10kW is R = E^2/P, R = 5.76 ohms, which is what you got. It's just a resistor and it doesn't have any kind of reactive component to it (that's worth mentioning ...), and it's not designed to conserve the amount of power the way a transformer does. What's happening here is that I = E/R and as you increase E from 240v to 480v you increase I in direct proportion to the amount you've increased E. Since you doubled E, you double I. Now, your 10kW oven is consuming 40kW because P = EI, and you've doubled both E and I, so you've quadrupled P.

Where you seem to be getting confused is that when it comes to Ohm's Law, you're not conserving the power through the system, which you do with a transformer. With a transformer,

EI = E'I'

where E and I are the input voltage and current and E' and I' are the output voltage and current (and they aren't exactly equal, but that's because nothing in life is fair or free).

 

[dbuckley]

what is confusing me is the fact that a typical home is full of resistive loads, yet the utility company greatly reduces current by raising the voltage on the grid supplying the load.

I don't get that.

I am happy that the utility can reduce their I2R losses in distribution by choosing to distribute (or supply) at a higher voltage. The losses go down by the square of the current, which makes it sensible to use the highest voltage practicable at all times. This is usually manifested by distributing or supplying at a higher standard voltage, 4160V rather than 480V.

But... all things being equal, increasing the voltage of supply to a consumer will increase current, as most loads are resistive (lights, furnace) or resistive like (eg motors).

Supply a bulb with more than the rated voltage and it doesn't use less current to maintain the stated output, it behaves exactly as Ohms Law states (even though R is not constant for a bulb, it varies by temperature), and takes more current, thus consuming more power, and being over-run, heading towards an early demise.

The only example I know of where increasing voltage results in the actual reduction of current consumed is for switch mode loads, such as computers and high frequency lighting, as these devices have power supplies that draw the required power from the supply, and as such are insensitive to supply voltage change, and will simply draw sufficient current to deliver the required power. In reality we are only changing supply voltage by a few volts here, you cant take a 240V furnace and wire it across 480V.

 

[LarryFine]

You're mixing some concepts in here.

Oops! Sorry, Julie, but so are you.

The POCO reduces the current by increasing the voltage -- that much is true. However, that's just so that P is a constant, while VD is reduced.

The only way you can reduce current by increasing voltage, while keeping power constant, is with an equipment change, such as changing transformer primary taps or replacing the transformer.

E = 120v, I = 100A, R = 0.1 ohms. At 100A a 0.1 ohm load results in a 10 volt drop. So, VD = 10v, and the voltage on the other end of the wire is 110v.
E = 240v, I = 50A, R = 0.1 ohms. At 50A a 0.1 ohm load results in a 5 volt drop. So, VD = 5, and the voltage on the other end of the wire is now 235v.

Yes, but the difference is that you have equipment manufactured for either 120v or 240v.

Now, you get to the question of loads and how E and I are related. When you have a fixed value of R, which you do for purely resistive, non-reactive loads, I = E/R. Increase E and I increases in direct proportion. That's Ohm's law. If you increase R, I decreases in inverse proportion. Again, Ohm's law.

Correctamundo, as is the rest of your post.


There is a difference between using equipment rated for different voltages, with the current inversely proportionate, in order to deliver a given amount of power; and varying the voltage to a given equipment, where the current varies proportionately (but not inversely) with the voltage.

Generally, a small increase in system voltage, with no equipment changes, will result in an overall increase in current, except for all-motor loads.

 

[tallgirl]

Oops! Sorry, Julie, but so are you.

The only way you can reduce current by increasing voltage, while keeping power constant, is with an equipment change, such as changing transformer primary taps or replacing the transformer.

Urph. You had to go confusing my otherwise very simplified post by getting all technical and factual on me.

Let him get the basics down before you start confusing him with transformer taps and reactive loads. Then we can make his head explode

 

[QES]

always use the grandold V=IR, this will solve everything! well almost everthing.

 

[AE-29]

[SIZE="5"Hey, did you guys consider that ohms law for[COLOR="DarkRed"] D.C[/COLOR] circuits refers to direct and inverse proportion to voltage,current and resistance. In A.C circuits impedance affects the circuit differently.