I don't intuitively believe that

[nhmllr]

if you have two resistors of the same ohmage, wired in parallel, that the resulting equivalent resistance for the circuit is half of either resistance

Shouldn't it be the same resistance?

I understand the equations, and how you obtain that value, but I just don't believe it. Can somebody explain, in terms of the physical reality of the circuit, why that would be the case?

Thanks

 

[A.T.]

Shouldn't it be the same resistance?

Can you "explain, in terms of the physical reality of the circuit" why it should be the same resistance?

 

[nhmllr]

Can you "explain, in terms of the physical reality of the circuit" why it should be the same resistance?

Shouldn't it be that half of the electrons go down one branch of the parallel wires, the other electrons go down the other wire, and both halves experience the same resistance once they're met up on the other side and continue?

 

[berkeman]

Shouldn't it be that half of the electrons go down one branch of the parallel wires, the other electrons go down the other wire, and both halves experience the same resistance once they're met up on the other side and continue?

No. Because you have the same voltage across both resistors, you get twice the total current that you would if you only had one resistor there. Then by Ohm's law, twice the current given the same voltage is the same thing as having half the resistance with the same voltage...

 

[ZapperZ]

if you have two resistors of the same ohmage, wired in parallel, that the resulting equivalent resistance for the circuit is half of either resistance

Shouldn't it be the same resistance?

I understand the equations, and how you obtain that value, but I just don't believe it. Can somebody explain, in terms of the physical reality of the circuit, why that would be the case?

Thanks

You DON'T have to "believe" anything. Take an Ohmmeter, and MEASURE them!

Zz.

 

[Muphrid]

Voltage = current x resistance.

Half the current goes through one resistor, half through the other. You get that. All this is saying, though, is that if half the current experiences the full resistance, then the full current experiences only half the resistance because the voltage must be the same either way. It's a consequence of the voltage drop being fixed.

 

[A.T.]

Shouldn't it be that half of the electrons go down one branch of the parallel wires, the other electrons go down the other wire,

Half of what? Why do you assume the total current is the same as with a single resitor? The parallel branches are independent of each other. Each having the same resistance R and put under the same voltage V has the same current I. So the total current is 2I.

 

[A.T.]

Once you have figured it out, try this one:

An electrical circuit has N nodes. Each node is connected to every other node by a resistor with the electrical resistance R (complete graph). What is the total resistance Rt between two nodes of the circuit?

 

[Dickfore]

According to 2nd Kirchhoff rule, the current flowing through each branch is the same as if only one resistor was conected across the same voltage. But, the current in the circuit, according to 1St Kirchhoff rule, is the sum of the currents in each branch, i.e. it is twice as large.

According to Ohm's law, the current in the circuit is inversely proportional to the resistance. Therefore, to get twice as big current, with the same voltage drop, you need half the resistance for the equivalent circuit. Q.E.D.

 

[Reptillian]

As per Oliver Heaviside's suggestion, I think of electronics in terms of fluid flowing through pipes when trying to understand conceptually. Resistance is analogous to a constriction in a pipe. If you add pipes in parallel, the fluid flows more easily.

I would post a link to a site that describes it all, but just google hydraulic analogy or electronic-hydraulic analogy you might get some insight.

(I can't post links until I have ten posts under my belt)

 

[luis20]

if you have two resistors of the same ohmage, wired in parallel, that the resulting equivalent resistance for the circuit is half of either resistance

Shouldn't it be the same resistance?

I understand the equations, and how you obtain that value, but I just don't believe it. Can somebody explain, in terms of the physical reality of the circuit, why that would be the case?

Thanks

I think I can explain: If you have two parallel wires, in each of them charges will be flowing with a rate of (V/R) coulombs/second. (remember I=V/R)

This means that you'll have 2*(V/R) coulombs/second flowing in those two wires. If you had only one wire, to be equal, the new resistance which I will call R* would have to allow a flow of 2*(V/R) coulombs/second. So I=V/R*=2*V/R, so R*=2/R.

Think of intensity I as coulombs/second

 

[yuiop]

Perhaps it might help to think of an analogy using water. Imagine you have a hose pipe connected to the bottom of a barrel. The head pressure of the water in the barrel can be thought of as the Voltage. If we allow the water to flow through the hose pipe the flow rate (litres per second) is equivalent to the electric current. If we connect another hose pipe in parallel to the barrel the current will double and using the relation V=IR, we can see that 2 hose pipes in parallel have half the resistance of a single hose pipe.

 

[Antiphon]

It's like Reptillian said- your bathtub would fill in half the time if you had two faucets.

You're thinking the flow in each faucet would be cut in half; that's called "a constant current source". But you have a voltage source which is like a steady pressure in the pipes. In that case two faucets runs water twice as fast.