Question

Three resistance of which two are equal when connected in series have an effective resistance of 30 ohm. When these three resistors are connected in parallel the effective resistance is 3 ohm. Find the values of individuals resistance.​

Answer 1

Answer:

here's the solution

Explanation:

let the resistors be R1,R2&R3

A\Q

R1+R2=30ohm

now,

1/R1+1/R2+1/R3=1/R

now solve this it will form the equation and then you will get the answer

Answer 2

When connected in series, R = R1 + R1 + R2 (two are equal)

R = 2R1 + R2

∴ 2R1 + R2 = 30 (R = 30Ω)

∴ 2R1 = 30 – R2

∴ R1 = 30 – R/2 …(i)

When connected in parallel, 1/R = 1/R1 + 1/R1 + 1/R2

Or 1/R = 1/(30 – R2)/2 + 1/(30 – R2)/2 + 1/R2 [from equation (i)]

∴ 1/R = 2/30 – R2 + 2/30 – R2 + 1/R2

1/3 = 4/(30 – R2) + 1/R2

∴ 1/3 = 4R2 + 30 – R2/R2(30 – R2) = 3R2 + 30/R2(30 – R2)

∴ R2(30 – R2)/3 = 9R2 + 90

30R2 – R22 = 9R2 + 90

R22 – 21R2 + 90 = 0

∴ R2 – 21R2 + 90 = 0

R2 – 6R2 – 15R2 + 90 = 0

∴ R2 = 6 or 15

∴ R1 = 30 – 6/2 or 30 – 15/2

R1 = 12Ω or 7.5Ω

The three resistance can have values 12Ω, 12Ω and 6Ω or 15Ω, 7.5Ω and 7.5Ω