Question

Two resistances R1 and R2, when connected in series give a resistance of 10Ω and a

resistance of 2.4Ω when connected in parallel. Find the value of R1 and R2.

Answer 1

series= r1 +r2= 10
parallel= r1r2/r1+r2 =2.4
r1r2/10=2.4
r1r2=2.4×10
r1r2=24
r2=24/r1
r1+r2=10
r1+24/r1=10
r1^2 +24=10r1
r1^2-10r1+24=0
r1^2-6r1-4r1+24=0
r1(r1-6)-4(r1-6)=0
(r1-6)(r1-4)=0
r1=4,6
r1+r2=10
if r1 =4
then 4 +r2=10
r2=10-4=6
r2=6
if r1=6
then 6+r2=10
r2=10-6=4
r2=4

Answer 2

in Series R1 + R2 = 10ohm-------eq.(i)

in parallell

R1*R2/ (R1+R2) = 2.4

R1*R2 = 2.4 (R1+R2)

R1 * R2 = 2.4 * 10 by eq.(i)

R1*R2 = 24------eq.ii

therefore applying arithmetic equation, we get

(r1+r2)^2 = r1^2 + r2^2 + 2*(r1*r2)

10*10 = r1^2 + r2^2 + 2 *24 ---by eq. (i) and (ii)

100 = r1^2 + r2^2 + 48

100-48 = r1^2 + r2^2 or 52 = r1^2 + r2^2

again

(r1-r2)^2 =[ r1^2 + r2^2] - [2*(r1*r2)]

putting values we get

(r1-r2)^2 = 52 - 48

(r1-r2)^2 = 4

(r1-r2) = 2

Now we have

r1 + r2 = 10

r1 - r2 = 2

adding both equation we get

2*r1 = 12

r1 = 6

there fore r2 = 10 - r1 [from equation (i)]

r2 = 10- 6 = 4/

your answer

r1 = 6

r2 =4