When R2 and R2 connected in series the net resistance in 6 ohm and in a parallel it is 3 ohm.find R1 and R2.
Answers
Answered by
1
R1+R2=6
1/R1+1/R2=1/3
R1=6-R2
1/6-R2+1/R2=1/3
6-R2+R2/6(R2)-R2^2
6/6R2-R2^2 =1/3
1/6R2-R2^2 =1/9
LET R2 BE a
-a^2 +6a - 9=0
a^2 - 6a+9=0
a^2 -3a - 3a +9=0
a(a-3)-3(a-3)
a-3=0 a=3
R2=3
R1=3
1/R1+1/R2=1/3
R1=6-R2
1/6-R2+1/R2=1/3
6-R2+R2/6(R2)-R2^2
6/6R2-R2^2 =1/3
1/6R2-R2^2 =1/9
LET R2 BE a
-a^2 +6a - 9=0
a^2 - 6a+9=0
a^2 -3a - 3a +9=0
a(a-3)-3(a-3)
a-3=0 a=3
R2=3
R1=3
Similar questions