basic question about resistance and Ohm's law

Hi, There is some basic stuff I don't understand about resistance. Legend holds it that resistance works by transforming power into heat. So why then, I a sk, would it result in decreased power usage by adding a resistor in a circ uit? In other words, let's say I have a light bulb in a 12 V circuit and th e current is 1 amp and thus the power usage is 12W. Thus the resistance is

12 Ohms, If I now add a 12 Ohms resistor to the circuit it doubles the resi stance and cuts the current and thus the power usage in half. But why? If t he lamp now uses half the power of the original circuit wouldn't the resis tor also use half the power by turning it into heat and once again the powe r usage would be 12W? Confused. Thank you,
Reply to
make1upper
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Resistor, capacitor, and inductor are three kinds of ideal linear components, that pass electrical current according to the voltage difference on two attachment points (terminals). Resistor is the only one that dissipates electrical power. All REAL components in an electrical circuit have some combination of these behaviors, i.e. are not IDEAL.

The key point here, is that 'adding a resistor in a circuit' can be many different things. The resistor in a battery/switch/lamp flashlight can be in series with the lamp, or in parallel, or in series with the battery, or in parallel, or in series with the switch, or in parallel.

If the switch is CLOSED and the resistance is in series with the lamp (which is itself a not-very-ideal resistor), it can turn the flashlight OFF if the resistor resistance is high compared to the lamp resistance. In that case, more resistance causes less power.

Reply to
whit3rd
12 Ohms, If I now add a 12 Ohms resistor to the circuit it doubles the resistance and cuts the current and thus the power usage in half. But why? If the lamp now uses half the power of the original circuit wouldn't the resistor also use half the power by turning it into heat and once again the power usage would be 12W? Confused. Thank you, ++++++++++++++++++++++++++++++++++++++++

Clue:

The bulb has 6 volts at 0.5 amps = 3 watts

Reply to
tm

Thank you for your replies. Maybe I didn't make myself clear enough: I understand Ohm's law and I understand what resistance means in a circuit. The problem I have understanding is that people say resistance turns into heat. If somebody told me that resistance only makes the pathway smaller fo r current (like a tap in a water pipe) and thus less power goes through the n I would completely understand everything about simple circuits and Ohm's law. But people say a resistor uses (dissipates) power. That's what I don't get. thanks

Reply to
make1upper

Legend? No.

Numerous careful measurements by smart physicists hold that resistance works by transforming power into heat.

Numerous burnt fingers by practical-minded engineers bear this out.

Because of ohms law. More resistance in the circuit means less current flow.

Leaving aside the fact that as the current goes down the lamp resistance also goes down, let's take your 12-ohm resistor by itself, and then your pair of 12-ohm resistors in series.

One 12-ohm resistor: With 12V applied, one amp flows. That's 12 watts being delivered from the supply, and its all being consumed (and turned into heat) by the resistor.

Two 12-ohm resistors, in series: With 12V applied, 1/2 amp flows. The voltage is evenly divided between the resistors, so each one has six volts applied.

The power supply is delivering 6W (12V * 1/2A). Each resistor is consuming 3W (6V * 1/2A). There are two resistors, whose total power is

6W, which equals the power delivered by the supply.

And once again, the first law of thermodynamics holds.

--
Tim Wescott 
Wescott Design Services 
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Reply to
Tim Wescott

Electrical power absorbed by a resistor comes out as heat.

P = E * I

P = power, in watts

E = voltage across the resistor, volts

I = current through the resistor, amps

You can do algebra things to get other equations, like

P = E^2 / R

So, if you slap a 1000 ohm resistor across a 9 volt battery, it dissipates 81 milliwatts. As heat. You could maybe barely feel 81 milliwatts. If you changed it to a 2000 ohm resistor, it would dissipate 40.5 milliwatts.

You have to evaluate specific situations and calculate what happens to each part. Statements like "resistance turns to heat" don't mean anything. Analogies don't help much either.

--
John Larkin                  Highland Technology Inc 
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Reply to
John Larkin

Right. Energy can not be destroyed. It can only be converted into smoke.

Reply to
tm

If you understand ohms law, it's a simple extension to understand power, and power dissipated as heat.

R = V / I and P = V x I

The analogy with a water tap and pipe isn't always very helpful. You're confusing power with pressure. Stick to pressure and flow, and power will follow.

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Mike Perkins 
Video Solutions Ltd 
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Reply to
Mike Perkins

Your terminology is a bit confusing. You are describing the circuit in words when this is a math concept. Yes, resistance turns current and voltage (which determines power) into heat. But to figure how much heat or power you have to do the math. Others have replied with some of the numbers.

You are describing the circuit in words but aren't analyzing it properly because you are making assumptions and aren't doing the math.

When you add the 12 ohm resistor you cut the current in half and the

*total* power is cut in half. But the light bulb dissipates a quarter of the original power (half the current times half the voltage) and the resistor dissipates a quarter of the original power totaling half.
--

Rick
Reply to
rickman

There is some basic stuff I don't understand about resistance. Legend holds it that resistance works by transforming power into heat. So why then, I ask, would it result in decreased power usage by adding a resistor in a circuit? In other words, let's say I have a light bulb in a 12 V circuit and the current is 1 amp and thus the power usage is 12W. Thus the resistance is

12 Ohms, If I now add a 12 Ohms resistor to the circuit it doubles the resistance and cuts the current and thus the power usage in half. But why?

** Because the current drawn from the battery is cut in half.

0.5 amps at 12V = 6 watts.

** That is where you are your mistaken.

The lamp now has BOTH half current AND half voltage since the resistor has

6V across it.

0.5 amps at 6V = 3 watts.

Resistors do two things, not one - they reduce current flow and reduce the available voltage in a circuit.

... Phil

Reply to
Phil Allison

Worse, the lightbulb is a nonlinear resistor.

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John Larkin                  Highland Technology Inc 
www.highlandtechnology.com   jlarkin at highlandtechnology dot com    
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Reply to
John Larkin

I never liked the water flow analogy. Flow is nonlinear on pressure drop in a pipe.

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John Larkin                  Highland Technology Inc 
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Reply to
John Larkin

Yes, that is definitely the problem here. The OP is confused because he knows the light bulb is non-linear.

--

Rick
Reply to
rickman

"rickman"

** I really doubt the OP is aware of any such thing.

His mistake was to not do the math - as you already said.

... Phil

Reply to
Phil Allison

You need to learn to recognize sarcasm. Sometimes the smiley isn't given.

--

Rick
Reply to
rickman

** OK , that was a tad subtle for this NG.

I often use " .... " at the end of such remarks in lieu of a smiley.

.... Phil

Reply to
Phil Allison

Good answers, good answers?. I guess my question is: why is Ohm's law correct and so different than comm on sense? If I knew nothing about electronics common sense would say that t he power consumption of one circuit illuminating a light bulb and a second circuit illuminating a light bulb (albeit dimmer) *and* heating up a resist or (of the same resistance as bulb) in series would be equal or even more.

my second question (not really related) is: so are resistors evil things? I t seem like nobody would want to waste energy by creating heat. Is there no way to "throttle" the flow of electrons without wasting power by turning i t into heat?

Along those lines, if I run a 6V circuit with a 6 Ohm bulb (pretending bulb s are linear resistors) it would consume 6W. If I want to power the same bu lb by a 12V battery I would have to add a 6Ohm resistor to have the bulb i lluminate at the same brightness. This circuit would consume 12W of power. That seems like an incedible waste of power. What am I missing here?

thanks alot for your input.

Reply to
make1upper

Good answers, good answers?. I guess my question is: why is Ohm's law correct and so different than common sense?

** You need to improve your common sense.

If I knew nothing about electronics common sense would say that the power consumption of one circuit illuminating a light bulb and a second circuit illuminating a light bulb (albeit dimmer) *and* heating up a resistor (of the same resistance as bulb) in series would be equal or even more.

** And if you considered the situation with battery drain, you might see just how dumb that is.

My second question (not really related) is: so are resistors evil things?

** Many engineers consider them evil.

Along those lines, if I run a 6V circuit with a 6 Ohm bulb (pretending bulbs are linear resistors) it would consume 6W. If I want to power the same bulb by a 12V battery I would have to add a 6Ohm resistor to have the bulb illuminate at the same brightness. This circuit would consume 12W of power. That seems like an incedible waste of power. What am I missing here?

** The bleeding obvious.

Use 6V lamps with only with 6V batteries.

Why do you think that lamps exist in all ratings from 1.5V up to 240V ?

... Phil

Reply to
Phil Allison

Assuming the lamp still has 12 ohms resistance when operated from 6 volts (It won't, lamps don't work like that, but that's immaterial to this example and would only make the arithmetic harder)

No, the lamp now only uses one quarter of the original it sees 6V and passes 0.5A for a total of 3W, the resistor sees another 3W.

for 3W+3W = 6W total energy used.

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Reply to
Jasen Betts

What do you think an electric heater does? It's a low resistance Nichrome resistor that turns current into heat..

Reply to
Michael A. Terrell

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