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Why does voltage remains same over parallel circuit. If a resistor is connected in the circuit some of the charge should be transformed into heat and make a lack of charge after the resistor (in my sense). So, what's the reason in it?

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Electric charge isn't transformed into heat by a resistor so your reasoning is based on a serious misunderstanding of the physics. – Alfred Centauri Oct 9 '13 at 2:24
What about light bulb? Its a sort of resistor...n it produces heat – Aritra Hazra Oct 9 '13 at 3:09
Electric energy is converted to heat, not electric charge. – Alfred Centauri Oct 9 '13 at 10:30

This kind of misconception can be cleared by the definition itself.

Voltage is the energy per free electron (which contributes to current flow in the conductor), whereas current is the rate of flow of free electrons across the conductor's cross-sectional area. In other words, current is the count of the stuff that passes through the cross-section within a given time period and voltage is what drives the stuff.

Charge is a conserved quantity. What you perceive as heat is the energy of the particles1 flowing at the drift speed (around a few millimeters per second). It's simply the voltage that's converted to heat. The free electrons can't smell around and divide accordingly based on resistance on each path of the circuit. It's just a random flow. They just go around and when the path-division is encountered, some go through one way and some go through the other.


To dig further inside, let's consider a parallel network like the one here (ABCDEFA). The battery (DC) maintains a potential difference (how much doesn't matter for now) which is far enough for the charges2 to start flowing. These charges encounter a junction B on the way. As previously told, there's no specific condition that reroutes the charges to some preferred direction. It's simply random. Hence, some follow the path BE, while the remaining go via CD to reach the battery.

Say the resistance of $R_2>R_1$. What would happen? The time taken by the charges to swim through $R_2$ is greater than it takes through $R_1$. So, a lot of charges can get out of $R_1$ within a specific time period, while the number is less in case of $R_2$. And, this happens within a few seconds once the potential difference is established and that's why we perceive that the current through $R_2$ is less than the one through $R_1$ (which is the reason why "current divides in parallel circuits).

Once the charges get out of the resistors, the electric field of the battery is enough to drive them mad (as the wire has relatively lower resistance). And, the charges get back their energy once again. This is the reason why we say voltage is the same in parallel circuits3.

1: I mentioned "particles" simply, because the free electrons aren't necessarily flowing at the drift velocity (which may lead to another misconception). They're always at relativistic speeds. The drift velocity is just a representation of their contribution to the current flow in the macro scale...

2: In a crude manner, I simply used "charges" because conventional current flow is from positive to negative, whereas the electron flow is the other way around (which is quite difficult for me to express). And by charges, I meant charged particles.

3: Also note that the voltage and current remain the same for resistors of the same resistance whether they're in parallel or series...

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The reason is that charge is conserved. Even if some of the energy is lost as heat, you will still end up with the same number of charge carriers in the wire.

Charge conservation is a physical law that states that the change in the amount of electric charge in any volume of space is exactly equal to the amount of charge flowing into the volume minus the amount of charge flowing out of the volume. In essence, charge conservation is an accounting relationship between the amount of charge in a region and the flow of charge into and out of that region.

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But how it's conserved when it's losing energy from the volume (in the form of heat) – Aritra Hazra Oct 9 '13 at 2:12
The charge is not destroyed when it loses energy. It just slows down. – mcodesmart Oct 9 '13 at 2:21

As the battery forces charge through the resistor, indeed that energy is converted into heat. An ideal battery has enough power so that it can supply the current without any voltage drop.

With a real battery, you have to make allowance for the battery's internal resistance. This resistance is inseries with the resistance in the circuit. Because of that, the current effectively flows through a voltage divider, and the voltage across the resistor is reduced. If the battery voltage is $V_B$ and it's internal resistance is $R_i$ then the voltage across the resistor $R$ is $$V=V_B*R/(R+R_i)$$ Unless $R_i=0$ the resulting voltage across the resistor is less than the battery voltage.

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A single resistor wired in parallel with voltage source like a battery may give confusing reasons for a novice, so a few other things need to be taken into account.

Any battery also has an internal resistance (it is like a resistor in series with the voltage source), and this resistance also varies with the state of charge of the battery, which is to say, the internal resistance gets to be a quite large value when the battery is "low". This internal resistance of a battery can be complicated to figure out; don't try connecting an ammeter directly across the terminals of a battery!

So, when the battery is fresh and the value of the external (test) resistor in parallel with it is in a range that will not cause it to overheat (which would abruptly change its value to something much larger than marked), it is easy to understand that a voltmeter connected across the resistor will hopefully be the same as if the resistor were not there, and the voltmeter is connected directly across the battery.

By the same token, adding additional resistances in parallel to the first one should also not materially change the value of the voltage measured in the first case unless the voltage source has been overloaded.

Does this help? Some of the other examples seemed not to take into account the idea that the question being asked seemed to involve a (single) resistor and a (single) voltage source, which should be thoroughly understood before adding more stuff in parallel, to say nothing of the characteristics of real voltage sources, particularly a battery.

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The definition of a perfect conductor is that there is no voltage drop along it's course. Of course nothing is perfect but copper is a very good conductor and it would be difficult to measure the voltage drop across relative short segments. In a parallel circuit the conductors on either side of the multiple resistive elements maintain nearly identical voltages. It is the current that gets divided. As others have pointed out, the charge is a conserved quantity (since it's just the numbers of electrons) so it gets "divided" between the various resistive elements and then gets recombined back at the other side so I_out is the same as I_in at the current source.

In the diagram offered by another poster, Waffle's Crazy Peanut, there is a "wire" from the start of R1 to the start of R2. In circuit theory that section of "wire" has no resistance and it's equivalent to both resistors being connected to the same point which implies that they have the same voltage at their positive sides.

Electrical engineers sometimes talk about across quantities and through quantities. Voltage is an across quantity while charge is a through quantity. You raise the issue of heat which is a measure of power generation. Heat from each resistor in a parallel circuit will be calculated as the product of the voltage (the same for each) and current (different and inversely proportional to the resistances.)

You should think of voltage as pressure and the charge as the fluid being pushed forward. If you turn off the faucet attached to a garden hose (decreasing pressure or by analogy the voltage) , the water doesn't decrease in amount; rather only the flow of charge (current) decreases.

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protected by Qmechanic Jan 7 '14 at 14:04

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