The equivalent resistances of two conductors are 9 Ω and 2 Ω when they are connected in series and parallel combinations respectively. Find the resistances of those two conductors.
OR
Find the equivalent resistance between A and B points of the following circuit.
For Series: R1 + R2 = 9 or, R1 = 9 – R2 — (i)
For Parallel: 1\over R_\text{parallel} = 1\over R_1 + 1\over R_2
or, 1\over 2 = R_1×R_2\over R_1 + R_2
or, 1\over 2 = (9 - R_2)×R_2\over 9
or, 9R2 – R22 = 18
or, R22 – 9R2 – 18 = 0
or, R22 – (6 + 3)R2 – 18 = 0
or, R2(R2 – 6) – 3(R2 – 6) = 0
or, (R2 – 6)(R2 – 3) = 0
or, R2 = 6Ω or 3 Ω
Put R2 in eq (i) we get: R1 = 6Ω or 3 Ω
The resistances of the two conductors are 3Ω and 6Ω.
OR
The 2 Ω and 3 Ω resistors are connected in parallel.
1\over \text{R}_{\text{Parallel}}= 1\over \text{R}_1 + 1\over \text{R}_2
1\over \text{R}_{\text{Parallel}}= 1\over 2 + 1\over 3
1\over \text{R}_{\text{Parallel}}= 3 + 2\over 6 = 5\over 6
or, Rparallel = 6\over 5 = 1.2 Ω
∴ Rtotal = Rparallel + R3 = 1 Ω + 1.2 Ω = 2.2 Ω
Related Questions
More WBBSE Hard level questions