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Hello everybody I have a question about resistance in Ohm's law, according to the formula "I = U/R" the less resistance you have, the more current you got and accordingly the more power you drain. So I don't get it, is this meaning the more load you got the less power you drain? That's a little counterintuitive, so I need a little advice from you guys to make this clear. Thank you in advance.

Just to make things clear.

  1. An electrical load is an electrical component or portion of a circuit that consumes (active) electricpower. E.g. solenoid, motor, lightbulb etc.
  2. The stronger load (solenoid), the more wire it has, the more resistance it has.
  3. The higher resistance in circuit, the lower current flow, and the power.
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  • \$\begingroup\$ Load refers to power ,P not resistance. So P=I²R=U²/R and yes I=U/R \$\endgroup\$ – Tony Stewart EE75 Jan 28 '18 at 0:51
  • \$\begingroup\$ If it makes you happier, you can consider the load's conductance, G, instead of its resistance. G is just the inverse of R, G = 1/R. Then Ohm's Law becomes I = V*G, and a larger conductance indicates a higher current if the applied voltage is fixed. \$\endgroup\$ – The Photon Jan 28 '18 at 1:15
  • \$\begingroup\$ So is that meaning the more powerful the load is (LED, Speaker, motor etc.), the less resistance it got ? \$\endgroup\$ – Zhirayr Jan 28 '18 at 1:39
  • \$\begingroup\$ If you are interested in understanding it mathematically and intuitively, I have given an answer here that might be of help: electronics.stackexchange.com/questions/543610/… \$\endgroup\$ – Ali Kavoosi Feb 4 at 1:54
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... according to the formula "I = U/R" the less resistance you have, the more current you got and accordingly the more power you drain.

Correct.

... is this meaning the more load you got the less power you drain? That's a little counterintuitive, ...

The smaller the resistance the more current will flow. (That should be fairly obvious - think of the water analogy: a bigger pipe has lower resistance and will have a larger flow for a given head of water.)

enter image description here

Figure 1. Water "current" and flow analogy. Note that the pressure (voltage) is the same in both cases. Source: Sparkfun.

The "load" is the load on the source, the energy supply or, in the case of the water analogy, the water supply or reservoir. The lower the resistance of the load the more current will flow.


Update after question updated:

  1. An electrical load is an electrical component or portion of a circuit that consumes (active) electricpower. E.g. solenoid, motor, lightbulb etc.

So far, so good.

  1. The stronger load (solenoid), the more wire it has, the more resistance it has.

Not necessarily. It's the ampere-turns that will determine the strength of the solenoid. If you double the number of turns but keep the wire gauge and voltage the same then the resistance will double and the current will halve. Double turns x half the current gives the same ampere-turns so no improvement.

To increase the ampere-turns you could:

  • Increase the wire size. (Lower resistance, more current -> higher ampere-turns.)
  • Increase the turns with the same wire size and increase the voltage in proportion to maintain the current. (Same current, more turns -> higher ampere-turns.)
  • Any combination of the above.
  1. The higher resistance in circuit, the lower current flow, and the power.

Correct.

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It is a bit counter intuitive. "More load" means a lower load resistance. More current flows into a smaller resistor. The device driving the load has to push harder than it does for a "small load" i.e., higher load resistance.

We can measure loading in units of Amperes. Placing a smaller resistor across the terminals of a power supply causes more Amperes to flow. This is an "increase in loading."

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  • \$\begingroup\$ So I assume the device that driving the load is the power supply, then I would like to know in what units we measure pushing of the device? \$\endgroup\$ – Zhirayr Jan 28 '18 at 0:22
  • \$\begingroup\$ @Zhirayr - I edited my answer. Does that make sense? \$\endgroup\$ – vofa Jan 28 '18 at 0:33
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the less resistance you have, the more current you got and accordingly the more power you drain

Well, not exactly: that's more power IF I*U goes up, but there's not necessarily any more current unless you think the voltage is a constant. Voltage and current are both variables, and only by examining the source character do we know how they vary.

the more load you got the less power you drain?

The more resistance, the less power from a voltage source. An ideal battery and light bulb and switch, with the switch 'off', i.e. high resistance, doesn't drain the battery.

Current can be the constant, too. A flashlight lamp filament (high resistance) heats greatly when the switch is closed, but the switch and wiring stay cool with the same current.

Ohm's law gives one equation in two unknowns, I and U. Power source character determines a second equation, and you can then solve two equations for the two unknowns, and compute the power I*U that is transferred.

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Much has been written about Ohm's Law in this forum and most of it is wrong. So I thought I would attempt to straight out some folk's thinking about the subject.

First of all, Ohm's Law is NOT R = V/I . That formula is the definition of resistance. A definition is not a law. Yet this mistake has been mindlessly propagated down through countless teachings, papers, courses and whatever. I can provide documentation for my assertion if challenged.

The true Ohm's law is a material property. It states that the resistance of a material does not change if the voltage used to measure the current varies. Copper follows Ohms Law, but a diode does not. I hope the above discussion clears things up.

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  • \$\begingroup\$ Regarding downvoting: It is a pitty and disappointing that there are some people who cannot accept that they had not the correct understanding in the past. Another misinterpretation of Ohms law: V=I*R. Of course, we are using this formula - but we must not interprete it as cause (current I) and effect (voltage V). A current cannot "produce" a voltage - in contrary: Each current needs a driving E-field (voltage) first. \$\endgroup\$ – LvW Feb 4 at 8:16
  • \$\begingroup\$ i I am unclear on where you stand on this subject. Do you believe that stating Ohm's Law as R = V/I us a misnomer or not? Or is R= V/I the definition of resistance and not a physical law? Which do you believe? \$\endgroup\$ – Ratch Feb 5 at 16:41
  • \$\begingroup\$ My comment was directed only to the guy who gave you a downvote (because I fully agree with your view) \$\endgroup\$ – LvW Feb 5 at 16:56
  • \$\begingroup\$ Thank you LvW. As a further clarification, consider this; velocity is defined as distance traveled per time. We do not have a law called some like, ah let's see, Newton's Law of Velocity, for instance, v=d/t , do we? So why does the the whole electrical industry call a simple definition a law? Inquiring minds would like to know. \$\endgroup\$ – Ratch Feb 6 at 17:57
  • \$\begingroup\$ Yes - unfortunately, in some contributions/discussion there is a mixture between correlation, causality, laws, definition and some practical rule of thumbs (example: Some people think that the equation Ic/Ib=10 for BJTs switchinng operation would be something like a law or even the definition for saturation.) \$\endgroup\$ – LvW Feb 7 at 9:33

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