When two resistors r1 and r2 are connected in parallel, the net resistancesatisfies the equation:1/r=1/r1+1/r2

Question

solve this equation for R1 in terms of R and R2.

a. R1=1/R-1/R2
b. R1=RR2/R2-R
c. R1=RR2/R-R2
d. R1=R-R2/RR2

Answer 1

1/R = 1/R1 + 1/R2

1/R1 = 1/R - 1/R2

= (R2 - R) / RR2

R1 = RR2 / (R2 - R)

So, b is the answer

Answer 2

Starting from

1/R = 1/R1 + 1/R2

we're trying to get something like R1 = ____.

As you can see, R1 is in the denominator. To make things simpler, lets just get rid of all the denominators by multiplying by the common denominator, (R*R1*R2). This gives:

(R*R1*R2)/R = (R*R1*R2)/R1 + (R*R1*R2)/R2

which simplifies (through cancellation) to:

R1*R2 = R*R2 + R*R1

Let's get all the R1's together on the left-hand side by subtracting R*R1 from both sides

R1*R2 - R*R1 = R*R2

Factor the R1 from the left-hand side (aka combine like terms):

(R2 - R)*R1 = R*R2

Then, divide by the coefficient of the R1 term, (R2 - R)

(R2 - R)*R1 / (R2 - R) = R*R2 / (R2 - R)

And, after cancellation on the left,

R1 = R*R2 / (R2 - R) (Answer B)

Answer 3

Its
1/R=1/R1+1/R2
So, 1/R1=1/R-1/R2
1/R1=(R2-R)/RR2
R1=RR2/R2-R
(b) R1=RR2/R2-R

Answer 4

The answer is b.
1/R=1/R(1) +1/R(2)
1/R={R(2)+(R1)} /R(1)R(2)
R(1)R(2)= {R(R(2)+R(1)}
R(1)R(2)=R*R(2)+R*R(1)
R(1)R(2)-RR(1)=RR(2)
R(1){R(2)-R} =RR(2)
R(1)= RR(2) / {R(2)-R}