If three resistors of 1 Ohm each connect in parallel to each other the resultant resistance is
- (a)1 Ohm
- (b)1⁄3 Ohm
- (c)3 Ohm
- (d)9 Ohm
Answer
Why
Correct — B, 1⁄3 Ohm. For resistors in parallel the reciprocals add: 1/R = 1/1 + 1/1 + 1/1 = 3, so R = 1/3 Ω. Combining equal resistors in parallel always gives a resistance below the smallest one — here three 1 Ω resistors give 1/3 Ω. (Quick check for n equal resistors in parallel: R = value ÷ n = 1 ÷ 3.)
Why the others are wrong
- (a)1 Ohm — 1 Ω would be the resistance of a single resistor; adding more in parallel must lower the total, not keep it the same.
- (c)3 Ohm — 3 Ω is the result for the three resistors in SERIES (they add), not in parallel.
- (d)9 Ohm — There is no combination of three 1 Ω resistors that gives 9 Ω; this simply multiplies incorrectly.
Concept
Resistors in parallel share the current; their reciprocals add, 1/R = 1/R₁ + 1/R₂ + 1/R₃, so the combined resistance is always less than the smallest individual resistor. Resistors in series carry the same current and simply add, R = R₁ + R₂ + R₃, giving a value larger than any single one. For n equal resistors R each, parallel gives R/n and series gives nR.
The examiner offers both the series answer (3 Ω) and the parallel answer (1/3 Ω) to catch a rushed reader. 'Parallel' means reciprocals add and the result drops below 1 Ω, so 1/3 Ω is correct.
Key facts
- In parallel the reciprocals add, 1/R = 1/R₁ + 1/R₂ + 1/R₃, so the total is smaller than the smallest resistor.
- In series the values add, R = R₁ + R₂ + R₃, so the total is larger than the largest resistor.
- n equal resistors R in parallel give R/n; in series they give nR.
- Three 1 Ω resistors: parallel = 1/3 Ω, series = 3 Ω.
Study next
Common traps
- Giving the series answer (3 Ω) when the question says parallel.
- Forgetting that a parallel combination is always less than the smallest resistor.
Asked as a direct series/parallel calculation, or by mixing series and parallel sections to find an equivalent resistance.
Related PYQs
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
Answer(c) 9.0 Ω
Same concept — combining resistors in series and parallel. That NDA item mixes both rules (5 + 7 in series, then parallel with 36); this one is the pure parallel case with three equal resistors.
Practice
- practice — not a real PYQ
Three resistors of 1 Ω each are connected in series. Their equivalent resistance is
- (a)1⁄3 Ω
- (b)1 Ω
- (c)3 Ω
- (d)9 Ω
Answer(c) 3 Ω — in series the resistances add: 1 + 1 + 1. - practice — not a real PYQ
Two resistors of 2 Ω each are connected in parallel. The equivalent resistance is
- (a)4 Ω
- (b)2 Ω
- (c)1 Ω
- (d)0.5 Ω
Answer(c) 1 Ω — two equal 2 Ω resistors in parallel give 2 ÷ 2 = 1 Ω.