Two resistances of 5.0 Ω and 7.0 Ω are connected in series and the combi- nation is connected in parallel with a resistance of 36.0 Ω. The equivalent resistance of the combination of three resistors is
24.0 Ω
12.0 Ω
9.0 Ω
6.0 Ω
Answer is Wrong!
Answer is Right!
This question was previously asked in
UPSC NDA-2 – 2024
R_series = R₁ + R₂ = 5.0 Ω + 7.0 Ω = 12.0 Ω.
Next, this series combination (with R_series = 12.0 Ω) is connected in parallel with a third resistance (R₃ = 36.0 Ω). For resistors in parallel, the reciprocal of the equivalent resistance (R_eq) is the sum of the reciprocals of the individual resistances:
1/R_eq = 1/R_series + 1/R₃
1/R_eq = 1/12.0 Ω + 1/36.0 Ω
To add the fractions, find a common denominator, which is 36.
1/R_eq = (3/36) + (1/36) = 4/36
1/R_eq = 1/9
R_eq = 9.0 Ω.