Sum of resistance R1 and R2 when connected in series is 9 and when connected parallel is 2 . Please help in finding our value of Ra and R2.
2006-11-14 19:17:44 · 5 answers · asked by sm 2 in Science & Mathematics ➔ Physics
in series
R1+R2=9
in parallel
R1R2/(R1+R2)=2
R1*R2=18
R1=18/R2
R2+18/R2= 9
R2^2 -9R2+18=0
(R2-6)(R2-3)=0
This is saying that in either R1=6 ohms and R2=3ohms or vice versa. It doesn't matter you get the same result.
2006-11-14 19:29:23 · answer #1 · answered by tbirdwrestler 2 · 0⤊ 0⤋
The equivalent resistance for resistors in series is just their sum.
R1 + R2 = 9
In parallel, the reciprocal of the equivalent resistance is equal to the sum of the reciprocals of the resistances.
1/R! + 1/R2 = 1/2
2 equations, 2 unknowns. I'll let you do the rest.
2006-11-15 03:23:42 · answer #2 · answered by Demiurge42 7 · 0⤊ 0⤋
R1+R2=9
R1R2/R1+R2=2
R1R2=2(R1+R2)
=2*9=18
thereforethe resistances are6 nd 3 ohms respectively
2006-11-15 03:28:55 · answer #3 · answered by raj 7 · 0⤊ 0⤋
We know that R1+R2=9, and (R1*R2)/(R1+R2)=2.
Hence (R1*R2)/9=2
So R1*R2=18
So R1=3 and R2=6. (or Vice versa)
2006-11-15 03:26:08 · answer #4 · answered by ravish2006 6 · 0⤊ 0⤋
R +r=9
Rr/(r+R)=2
Rr=18
R(9-R)=18
R^2-9R+18=0
so R=6 or 3
so resistances are 6ohm and 3ohm
2006-11-15 03:29:17 · answer #5 · answered by yog 2 · 0⤊ 0⤋