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QUE☞ Two resistances when connected in parallel give resultant value of 2 Ω, when connected in series the value becomes 9 Ω. Calculate the value of each resistance.
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QUE☞ Two resistances when connected in parallel give resultant value of 2 Ω, when connected in series the value becomes 9 Ω. Calculate the value of each resistance.
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ikaptjcqbe
Answers & Comments
Answer:
Let r 1 andr 2 be the resistance
if it connected in parallel R eq (parallel)=
r 1×r 2÷r 1+ r 2 =2Ω...eq(1)
if it connected in series R eq (series)=r 1 +r 2
=9Ω..eq(2)
put the value of r 1 +r 2 andr 1 fromeq(2)ineq(1)
= r1× (9−r 1 ) =2Ω
(r 1 ) square −9r 1 +18=0
from this we get r 1 = 6Ω,3Ω
corrosponding to 6Ω we get r 2 =3Ω and 3Ωr 2 =6Ω
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Verified answer
Answer:
Explanation:
The same resistors when connected in series gives an equivalent resistance of 9 Ω. Let, R1 and R2 be two resistances. R1 = 6 Ω. and R2 = 9 – 6 = 3 Ω.
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