Two resistances of 5.0 Ω and 7.0 Ω are connected in series and the combination is connected in parallel with a resistance of 36.0 Ω. The equivalent resistance of the combination of three resistors is
- (a)24.0 Ω
- (b)12.0 Ω
- (c)9.0 Ω
- (d)6.0 Ω
Correct — C, 9.0 Ω. First add the two series resistors: 5.0 + 7.0 = 12.0 Ω (series resistances simply add). This 12.0 Ω branch is then in parallel with 36.0 Ω, so use the product-over-sum rule: (12.0 x 36.0) / (12.0 + 36.0) = 432 / 48 = 9.0 Ω. A useful check: the equivalent of a parallel combination is always LESS than the smallest resistor in it (here less than 12.0 Ω), and 9.0 Ω satisfies that.
- (a)24.0 Ω — 24.0 Ω comes from wrongly averaging the two parallel branches, (12 + 36)/2 = 24. Averaging is valid only for equal resistors; more decisively, a parallel result can never exceed the smallest branch (12 Ω), so 24 Ω is impossible.
- (b)12.0 Ω — 12.0 Ω is only the series sub-total (5 + 7). It ignores the second step — putting that 12 Ω in parallel with 36 Ω — which must pull the value below 12 Ω.
- (d)6.0 Ω — 6.0 Ω is a mis-computed parallel value (e.g. dividing 432 by 72 instead of by the correct sum 48, or halving the 12 Ω branch as if the two were equal). The correct denominator is 12 + 36 = 48, giving 9.0 Ω, not 6.0 Ω.
Resistors in SERIES add directly (R = R1 + R2 + ...). Resistors in PARALLEL combine by reciprocals (1/R = 1/R1 + 1/R2 + ...); for exactly two, the product-over-sum shortcut R = R1R2/(R1+R2) is fastest. Mixed networks are solved by collapsing the innermost series/parallel groups one step at a time.
The trap is to stop after the series step (getting 12 Ω, option b) or to average the branches (getting 24 Ω, option a). Do it in two stages: series first (5+7 = 12), then parallel (12 with 36). Knowing that a parallel equivalent is always smaller than its smallest member lets you eliminate 24 Ω and 12 Ω on inspection before you even compute.
- Series: resistances add — 5.0 + 7.0 = 12.0 Ω.
- Two-resistor parallel shortcut: R = R1R2/(R1+R2) = (12x36)/48 = 9.0 Ω.
- A parallel combination is always LESS than its smallest resistor (here < 12 Ω).
- Parallel lowers resistance because it adds more paths for current; series raises it by lengthening the single path.
Collapse the series pair first, then combine that branch in parallel with 36 Ω.
- Stopping after the series step and answering 12 Ω.
- Averaging the two parallel branches (only valid when they are equal).
- Forgetting that a parallel equivalent is always less than the smallest resistor.
NDA/UPSC give a small mixed series-parallel network and ask for the equivalent resistance — always identify series and parallel groups, then reduce in stages.
No directly related past PYQ was found.
- practice — not a real PYQ
Three resistors of 6 Ω, 6 Ω and 6 Ω are all connected in parallel. Their equivalent resistance is
- (a)18 Ω
- (b)6 Ω
- (c)3 Ω
- (d)2 Ω
Answer(d) 2 Ω — for n equal resistors in parallel, R = R1/n = 6/3 = 2 Ω.
- practice — not a real PYQ
A 4 Ω and a 12 Ω resistor are connected in parallel. The equivalent resistance is
- (a)16 Ω
- (b)8 Ω
- (c)3 Ω
- (d)48 Ω
Answer(c) 3 Ω — (4x12)/(4+12) = 48/16 = 3 Ω, which is less than the smaller resistor (4 Ω).