PDA

View Full Version : Beginner Question Regarding Paralleled Resistors.



Jim McCorison
01-31-2005, 12:24 AM
I've warned some of you that while I'm a grizzled old coder, I am somewhat of a novice at the hardware side of things. I am now trying to "fill in the blanks". As such I am starting at the basics and trying to better understand what I've simply excepted as fact before. One thing I've been reviewing is basic analog circuitry. Really basic. Like resistor circuits.

The resistance through a pair of paralleled resistors is defined as (R1 * R2) / (R1 + R2). If R1 is 1K and R2 is 2K, the circuit resistance is 666.67 ohms. Why do you have a resistance that is less than either of the resistors? http://forums.parallax.com/images/smilies/confused.gif

Jim

MacGeek117
01-31-2005, 12:29 AM
Because the current flowng through the circuit is greater than with one resistor.
bugg

▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
So many projects, so little time.

Jon Williams
01-31-2005, 12:38 AM
Put another way, adding parallel resistors increases the pathways for current to flow -- hence the circuit resistance is lower.· And note that in a parallel circuit, the overall resistance will always be something less than the smallest parallel resistor.

For parallel resistors you can use this formula with any number:

R = 1 / (1/R1 + 1/R2 ... 1/Rn)

▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
Jon Williams
Applications Engineer, Parallax
Dallas, TX· USA

Paul Baker
01-31-2005, 12:47 AM
Its easier to think in terms of conductance rather than resistance, each resistor when presented a voltage difference between its two terminals will conduct a certain amount of current, when you place more than one resistor in parallel each will conduct its own current as dictated by I=V/R since the total conductance is greater the equivalent resistance is lowered. Conductance is measured in Seimens (S), S=1/R. The conductance of resistors in parallel is St = S1 + S2 + S3 ... Translating that back to resistance 1/Rt = 1/R1 + 1/R2 + 1/R3 ... or Rt = 1/(1/R1 + 1/R2 + 1/R3 ...) which for two resistors in parallel is mathmatically equivalent to the equation you quoted.

Haha seems Jon and I were writing the same answer at the same time.

Jim McCorison
01-31-2005, 04:32 AM
Ok. That's pretty obvious when you stop to think about it. But somehow it escaped me.

So to paraphrase, each individual branch allows a certain amount of current to flow. When you add together the current flowing through each resister, the end amount is greater than what the lowest resistance would have allowed, thus the effective resistance of the circuit is less.

Thanks guys,
Jim