Voltage divider - stupid question

[nForce]

Ok, this is a classic question about a voltage divider:

Diagram:



If voltage on R5 and R6 are the same, why don't we just calculate for R5: R5/(R4+R5) which is equal to R6?

[jeroen79]

R5 and R6 are connected in parallel.
You need to calculate their equivalent resistance before you can work out the voltage divider.

[nForce]

Yes I know that, But why should we calculate equivalent resistance, we can just calculate for one resistor. Can you prove it?

[jancumps]

hi, you can’t calculate for one resistor because the other one influences it.

the value over the two is indeed the same, but it’s the result of their combined parallel value. not on the individual value of one of them.

so you need to first work out the value of the two in parallel. then use that result to calculate the divided voltage. that divided voltage will appear accross both resistors equally.

[zvonex66x]

Hi, for the first thing, u need to understand that R5 and R6 is in parallel, so togerher gives u abouth 500ohm, and then u can calculate with R4

[rstofer]

You can't assign a voltage to the top of the parallel resistors until after you figure out how much R4 is dropping and you can't get to that until you know how much current is flowing in the entire circuit.  It should be clear that the voltage drop across R4 will be different if the bottom resistor is 1000 ohms or 500 ohms.

So, you calculate the equivalent resistance and then you can use the voltage divider equation
Vout = Vin * (Req/(R4+Req))

[Wimberleytech]

To avoid jail, abide by the law.

Law #1: Ohm's Law
Law #2: Kirchoff's Law
Law #3: Abide by Law #1 and Law #2

Regards,
Texas Ranger

[bson]

As mentioned above, the voltage follows Ohm's law, so depends on the current: U=R*I, and the currents in your circuit depends on the resistances.  In other words, adding the parallel R6 will make more current flow through R4, but not through R5.  (KCL)  The current through R5 will in fact decrease.  Hence the voltage will increase over R4 and decrease over R5, showing that the actual branch resistances matter in a divider.

More generally, in this case you have a current divider.  Ohm's law is the law, and when you change one of U,R,I at least one of the others have to change as well.  In a fixed-resistance device like a resistor, it's not going to be R, at least not appreciably.

[Brumby]

If voltage on R5 and R6 are the same, why don't we just calculate for R5: R5/(R4+R5) which is equal to R6?

Yes, the voltage across R5 and R6 are the same - but you are assuming the voltage across R4 is the same as well.

It is not.

[Brumby]

Yes I know that, But why should we calculate equivalent resistance, we can just calculate for one resistor.

That's the point - you can't just calculate for one resistor.

Here's a challenge:  Calculate the way you want to and then calculate using the equivalent resistance as everyone else here is saying.  You will get two different answers - and only one will be correct.

Wanna guess which one?

Can you prove it?

Build it.  Measure it.

[rstofer]

Yes I know that, But why should we calculate equivalent resistance, we can just calculate for one resistor. Can you prove it?

Sure!

Assume you have a 10V battery, just to pick a number.

If you only have one bottom resistor, the total resistance is 2k and the current is 5 mA (10V/2K).  So, 5 mA times the bottom resistor of 1k gives 5v - exactly half the source voltage.  Just exactly what we expect.  The top resistor also drops 5V.  That's going to be important.  The voltage drops MUST add up to the voltage gains (battery).

Now solve for the case where the equivalent resistor is R5||R6 or 500 Ohms.  Now the total resistance is 1500 Ohms (1k top resistor + 500 Ohm equivalent resistor) and the current through R4 is 10V / 1500 Ohms or 6.666 mA.  Fine, take that 6.666 mA through R4 (1000 Ohms) and it drops 6.666V.  This means that the voltage at the top of R5||R6 is 3.333V (10V - 6.666V).  Each of the two 1k resistors will have 3.333V / 1k or 3.333 mA of current.  Two parallel paths of 3.333 mA add up to the original 6.666 mA coming into the junction of the 3 resistors.  6.666 mA coming in from R4 and twice 3.333 mA going out through R5 and R6.

Your proposed solution of just using one bottom resistor is just plain wrong.