SOLUTION: The combined resistance of two resistors R1 and R2 in a parallel circuit is given by the formula RT is lowered = 1 over 1 over R 1 (one is lowered) plus 1 over R2(two

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: The combined resistance of two resistors R1 and R2 in a parallel circuit is given by the formula RT is lowered = 1 over 1 over R 1 (one is lowered) plus 1 over R2(two       Log On


   

Question 83266: The combined resistance of two resistors R1 and R2 in a parallel circuit is given by the formula
RT is lowered =
1
over
1
over R 1 (one is lowered)
plus
1
over
R2(two is lowered)
simply the formula
Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
The formula for the total resistance R%5Bt%5D of two resistors R%5B1%5D and R%5B2%5D is given
by:
.
1%2FR%5Bt%5D+=+1%2FR%5B1%5D+%2B+1%2FR%5B2%5D
.
Let's begin by working on the right side. A common denominator for the two fractions
on the right side is the product of the two denominators or R%5B1%5D times R%5B2%5D. Suppose
you multiplied both of the two fractions on the right side by R%5B1%5D%2AR%5B2%5D%2F%28R%5B1%5D%2AR%5B2%5D%29.
This is equivalent to multiplying both of the fractions by 1 because the numerator of
the multiplier is the same as the denominator. So by doing this multiplication you are not
changing the value of the fractions ... just their form. The multiplication results in
the fractions becoming:
.
First fraction multiplied ---> %281%2FR%5B1%5D%29%2A%28R%5B1%5D%2AR%5B2%5D%2F%28R%5B1%5D%2AR%5B2%5D%29%29
.
But note that one R%5B1%5D in the numerator cancels with an R%5B1%5D in the denominator
so the first fraction becomes:
.

.
Retaining the common denominator R%5B1%5D%2AR%5B2%5D you are left with the first fraction becoming:
.
R%5B2%5D%2F%28R%5B1%5D%2AR%5B2%5D%29
.
Next work on the second fraction ----> %281%2FR%5B2%5D%29%2A%28R%5B1%5D%2AR%5B2%5D%2F%28R%5B1%5D%2AR%5B2%5D%29%29
.
But note that one R%5B2%5D in the numerator cancels with an R%5B2%5D in the denominator
so the second fraction becomes:
.

.
Note that you have retained the common denominator R%5B1%5D%2AR%5B2%5D and you are left with
the second fraction becoming:
.
R%5B1%5D%2F%28R%5B1%5D%2AR%5B2%5D%29
.
Substituting these two transformed (but unchanged in value) fractions back into the original
given equation results in:
.
1%2FR%5Bt%5D+=+R%5B2%5D%2F%28R%5B1%5D%2AR%5B2%5D%29%2B+R%5B1%5D%2F%28R%5B1%5D%2AR%5B2%5D%29
.
Now notice that the two fractions on the right side have a common denominator. Therefore,
these two fractions can be added by adding their numerators and placing that sum over the
common denominator to make the equation become:
.
1%2FR%5Bt%5D+=+%28%28R%5B2%5D+%2B+R%5B1%5D%29%2F%28R%5B1%5D%2AR%5B2%5D%29%29
.
Now all you have to recognize is that if you invert the two sides of the equation, you
do not change the equality. In other words, the inverted left side of the equation becomes:
.
R%5Bt%5D%2F1 which is just R%5Bt%5D
.
and the inverted right side of the equation becomes:
.
R%5B1%5D%2AR%5B2%5D%2F%28R%5B2%5D+%2B+R%5B1%5D%29
.
So the simplified equation is:
.
R%5Bt%5D+=+R%5B1%5D%2AR%5B2%5D%2F%28R%5B2%5D+%2B+R%5B1%5D%29
.
And the sum in the denominator is usually written in reverse order so the simplified
form becomes:
.
R%5Bt%5D+=+R%5B1%5D%2AR%5B2%5D%2F%28R%5B1%5D+%2B+R%5B2%5D%29
.
This formula is usually described as "two resistors in a parallel circuit can be replaced
by a single resistor having a value equal to the product of the two resistors divided
by their sum."
.
Hope you can see your way through this explanation and that it helps you to understand
a little more about both math and electronics.