# How do we explain powering a load without converting any current?

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#### Mark Flint

Joined Jun 11, 2017
145
Take a simple circuits like a 12v battery and a 5 amp filament bulb.

If we place meters before the bulb and after the bulb we see that current is 5 amps on both the positive and negative sides, while the voltage drops to a few millivolts.

Are we not taught that our electrical power is converted to light and heat in a light bulb? But if the current on the negative side of the circuit is identical to that on the positive side, what exactly was converted to light and heat?

If power = volts x amps and no amps were converted does that mean the bulb was powered completely by voltage alone? If not - what else is powering the bulb?

Thanks.

#### BobTPH

Joined Jun 5, 2013
8,104
Energy,

Bob

#### BobTPH

Joined Jun 5, 2013
8,104
Think if a hydroelectric dam. Does less water flow out of the dam than flows in? Of course not. The water just has less potential energy. The same is true of the electrons flowing out of and into the battery.

Bob

#### Irving

Joined Jan 30, 2016
3,548
Its the potential difference (PD in volts) across the bulb * the current through the bulb (in Amps) = energy transferred as heat/light (in Watts)

The PD is directly analogous to the potential energy lost by the water flowing over the dam in the example above.

#### Papabravo

Joined Feb 24, 2006
20,603
Ohm's law allows us to calculate power delivered to a load in three different ways. All three ways give the same answer. If they didn't there would be a serious imbalance in the "force".
$$P\;=\;VI\;=\;I^2R\;=\;\cfrac{V^2}{R}$$

#### Jim@HiTek

Joined Jul 30, 2017
59
It's magic. Duh.

#### ElectricSpidey

Joined Dec 2, 2017
2,631
Don't mean to be pedantic, but isn't that Watt's law?

(P = EI)

#### Irving

Joined Jan 30, 2016
3,548
No, its named after Watt but he didn't formulate it.

Wikipedia says:
The watt is named after the Scottish inventor James Watt.[5] This unit name was proposed initially by C. William Siemens in August 1882 in his President's Address to the Fifty-Second Congress of the British Association for the Advancement of Science.[6] Noting that units in the practical system of units were named after leading physicists, Siemens proposed that watt might be an appropriate name for a unit of power.[7] Siemens defined the unit consistently within the then-existing system of practical units as "the power conveyed by a current of an Ampère through the difference of potential of a Volt".[8]

#### nsaspook

Joined Aug 27, 2009
12,296
Take a simple circuits like a 12v battery and a 5 amp filament bulb.

If we place meters before the bulb and after the bulb we see that current is 5 amps on both the positive and negative sides, while the voltage drops to a few millivolts.

Are we not taught that our electrical power is converted to light and heat in a light bulb? But if the current on the negative side of the circuit is identical to that on the positive side, what exactly was converted to light and heat?

If power = volts x amps and no amps were converted does that mean the bulb was powered completely by voltage alone? If not - what else is powering the bulb?

Thanks.
Current (movement of charge) is not electrical energy. You can think of current like the links of a bicycle chain in this analogy. They just circulate round and round from energy source to energy sink as part of a energy transportation system chain using Tension.

#### crutschow

Joined Mar 14, 2008
33,352
If we place meters before the bulb and after the bulb we see that current is 5 amps on both the positive and negative sides, while the voltage drops to a few millivolts.
Nope.
Where do you measure a few mV?
The voltage across the bulb is 12V and the current through it is 5A.
So the power is 5A x 12V = 72W.

#### ElectricSpidey

Joined Dec 2, 2017
2,631
"No, its named after Watt but he didn't formulate it."

Nor did Ohm.

It's the name of the law.

#### Mark Flint

Joined Jun 11, 2017
145
So - just to be clear - nothing is getting converted, ie, from one form of energy to another - such as from chemical (battery) to light/heat. I'm thinking about the laws of thermodynamics - doesn't something need to be converted to end up being heat?

#### Irving

Joined Jan 30, 2016
3,548
while the voltage drops to a few millivolts.
This is a common misunderstanding.. treating voltage as if it is a lump of a substance rather than the difference between two (energy) levels. The analogy with potential energy and 'height' is a useful aid to understanding

#### Irving

Joined Jan 30, 2016
3,548
So - just to be clear - nothing is getting converted, ie, from one form of energy to another - such as from chemical (battery) to light/heat. I'm thinking about the laws of thermodynamics - doesn't something need to be converted to end up being heat?
Yes energy is converted from one type to another, but never lost or destroyed.

If I lift a weight up then let it fall i have used energy in my arm to give potential energy to the weight. when it falls it converts the potential energy to kinetic energy of motion...

similarly, a battery stores chemical energy. when a circuit is completed the potential difference resulting from that chemical energy drives an electrical current through the circuit. Components in the circuit convert that energy to heat, light, motion, sound, etc and in each the energy converted is the potential difference across them * the current through them

#### BobTPH

Joined Jun 5, 2013
8,104
nothing is getting converted
Nobody said nothing is converted.

Bob

#### nsaspook

Joined Aug 27, 2009
12,296
So - just to be clear - nothing is getting converted, ie, from one form of energy to another - such as from chemical (battery) to light/heat. I'm thinking about the laws of thermodynamics - doesn't something need to be converted to end up being heat?
If you mean the charge carrier thermodynamics, the energy they carry as current in the circuit energy transmission (good conductor/low acceleration) path is the charge carrier mass KE (this is the wiring resistive component loss). The electrical energy PE is in the fields here. At the light bulb load the higher resistance current path causes the conversion of the EM field energy PE to KE as charge carriers accelerate in the load electric potential, causing increasing collisions that result in thermal heat/ EM light energy.

#### Mark Flint

Joined Jun 11, 2017
145
Nope.
Where do you measure a few mV?
The voltage across the bulb is 12V and the current through it is 5A.
So the power is 5A x 12V = 72W.
I’m just going from memory of seeing a few mV on the negative side, perhaps I was wrong or my meter not accurate. But if there was zero potential difference between the load and the negative terminal does that mean that current can flow where there is no voltage difference?

#### Irving

Joined Jan 30, 2016
3,548
What you saw maybe was the voltage drop across the wire between bulb and battery -ve terminal. The wire has resistance and at 5A a few mV isn't unreasonable....

Unless the wire is a superconductor there will always be a PD across it... as there would be across the one from batt+ to bulb...

Neither has any bearing on the issue... other than to marginally reduce the PD across the bulb

#### nsaspook

Joined Aug 27, 2009
12,296
I’m just going from memory of seeing a few mV on the negative side, perhaps I was wrong or my meter not accurate. But if there was zero potential difference between the load and the negative terminal does that mean that current can flow where there is no voltage difference?
Sure, you need a electric field but it's a little complicated.
Let's be clear on current first.

You need an electric field to drive current. So how is an electric field established and maintained in the interior of a wire in a circuit? And how long does it take an electron in a wire to travel from my light switch to the bulb? Are electrons fast, or are they slow?

#### Mark Flint

Joined Jun 11, 2017
145
Sure, you need a electric field but it's a little complicated.
Let's be clear on current first.

You need an electric field to drive current.
I watched the first video, thanks, but haven’t finished the second yet, but it seems you still need a potentially difference for a current, and I’m still trying to understand how current returns to the battery on the negative side of the load.

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