Two resistances when connected in parallel give resultant value of $2 o h m,$ when connected in series the value becomes $9 o h m .$ Calculate the value of each resistance.

Open in App

Solution

Let $r_{1} a n d r_{2}$ be the resistance
if it connected in parallel $R_{e q} (p a r a l l e l) = \frac{r_{1} r_{2}}{r_{1} + r_{2}} = 2 Ω . . . e q (1)$
if it connected in series $R_{e q} (s e r i e s) = r_{1} + r_{2} = 9 Ω . . e q (2)$
put the value of $r_{1} + r_{2} a n d r_{1} f r o m e q (2) i n e q (1)$
$= \frac{r_{1} (9 - r_{1})}{9} = 2 Ω$
$(r_{1})^{2} - 9 r_{1} + 18 = 0$
from this we get $r_{1} = 6 Ω, 3 Ω$
corrosponding to $6 Ω$ we get $r_{2} = 3 Ω$ and $3 Ω r_{2} = 6 Ω$

Suggest Corrections

Similar questions

When n resistances each of value R are connected in parallel, then resultant resistance is x. When these n resistances are connected in series, total resistance is

Two resistances $R_{1}$ and $R_{2}$ give an equivalent resistance of 4.5 $Ω$ when connected in series and 1 $Ω$ when connected in parallel. Then the value of resistances are___________.

Join BYJU'S Learning Program

Two resistances when connected in parallel give resultant value of 2 ohm, when connected in series the value becomes 9 ohm. Calculate the value of each resistance.

Let r1andr2 be the resistanceif it connected in parallel Req(parallel)=r1r2r1+r2=2Ω...eq(1)if it connected in series Req(series)=r1+r2=9Ω..eq(2)put the value of r1+r2andr1fromeq(2)ineq(1) =r1(9−r1)9=2Ω(r1)2−9r1+18=0 from this we get r1=6Ω,3Ωcorrosponding to 6Ω we get r2=3Ω and 3Ωr2=6Ω

Two resistances when connected in parallel give resultant value of $2 o h m,$ when connected in series the value becomes $9 o h m .$ Calculate the value of each resistance.