A wire of resistance R is cut into four equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R′, then the ratio R′⁄R is:
- (a)1⁄16
- (b)1⁄4
- (c)4
- (d)16
Correct — A, 1⁄16. Cutting acts on the length and paralleling acts on the count, and both work in the same direction. A uniform wire has R = ρL/A, so resistance is proportional to length: four equal pieces each have resistance R/4. Four identical resistances of R/4 in parallel give an equivalent of (R/4) ÷ 4 = R/16. Hence R′/R = 1/16. The general rule is worth memorising — cut a wire into n equal parts and connect them all in parallel and the resistance falls by a factor of n², here 4² = 16.
- (b)1⁄4 — This is the result of doing only one of the two operations — either cutting the wire into four and stopping, or paralleling four whole wires of resistance R. The question does both, so the factor is applied twice.
- (c)4 — The ratio inverted. R′ is smaller than R, because putting conductors in parallel always gives an equivalent resistance smaller than the smallest branch.
- (d)16 — The right number the wrong way up. It is R/R′ rather than R′/R, and it would describe a combination in series, not in parallel.
Resistance depends on the geometry of the conductor as well as on the material: R = ρL/A, where ρ is resistivity, L is length and A is cross-sectional area. Cutting a uniform wire does not change ρ or A, so each piece keeps its share of the original resistance in proportion to its length. Combining resistances in parallel adds their conductances, so n equal resistances r give r/n.
Do the two steps separately and the arithmetic never goes wrong. Step one: four equal pieces, so each is R/4. Step two: four equal branches in parallel, so 1/R′ = 4 ÷ (R/4) = 16/R, giving R′ = R/16. The n² rule is what makes this a five-second question once you have seen it. Cut into two and parallel them and the resistance is R/4; into five, R/25; into ten, R/100. It is also worth noticing that the resistivity ρ is a property of the material and does not appear in the answer at all — cutting a copper wire and cutting a nichrome wire produce the same ratio, which is why the question can be answered without a single number. The reverse operation is the same rule in mirror image: joining n identical wires end to end multiplies the resistance by n.
- R = ρL/A — for a uniform wire of fixed cross-section, resistance is proportional to length.
- Cutting a wire of resistance R into n equal parts gives each part a resistance R/n.
- n equal resistances r in parallel give an equivalent resistance r/n.
- Net effect of cutting into n parts and paralleling all of them: resistance falls by a factor of n². Here n = 4, so R′ = R/16.
- Any parallel combination has an equivalent resistance smaller than its smallest branch; any series combination is larger than its largest.
Cutting divides by n, paralleling divides by n again — the total factor is n².
- Applying only one of the two operations and answering 1/4.
- Reporting R/R′ when the question asks for R′/R.
- Assuming the material or the original resistance value is needed — the ratio is purely geometric.
Asked as a ratio with no numbers at all, so the whole item is the n² rule for cutting and paralleling.
An electric wire of resistance 50 ohm is cut into five equal wires. These wires are then connected in parallel. What is the equivalent resistance of this combination?
- (a) 2 ohm
- (b) 10 ohm
- (c) 0·5 ohm
- (d) 5 ohm
Answer(a) 2 ohm
The identical question with numbers instead of a ratio and five pieces instead of four: 50 ÷ 5² = 2 ohm. Solving both makes the n² rule stick better than either alone.
A metallic wire having resistance of 20 Ω is cut into two equal parts in length. These parts are then connected in parallel. The resistance of this parallel combination is equal to
- (a) 20 Ω
- (b) 10 Ω
- (c) 5 Ω
- (d) 15 Ω
Answer(c) 5 Ω
The smallest case of the same rule, where 20 ÷ 2² = 5 Ω. Option (b) is there for anyone who cuts the wire and forgets to parallel the halves.
Three resistors of resistances 11 Ω, 22 Ω and 33 Ω are connected in parallel. Their equivalent resistance is equal to
- (a) 66 Ω
- (b) 22 Ω
- (c) 12 Ω
- (d) 6 Ω
Answer(d) 6 Ω
The parallel rule when the branches are unequal, which is the general case behind the shortcut used here. Adding reciprocals gives 1/11 + 1/22 + 1/33 = 11/66, so the equivalent is 6 Ω — smaller than every branch, exactly as a parallel combination must be.
- practice — not a real PYQ
A wire of resistance 36 Ω is cut into three equal parts and the parts are joined in parallel. The equivalent resistance is
- (a)4 Ω
- (b)12 Ω
- (c)9 Ω
- (d)108 Ω
Answer(a) 4 Ω — each part is 36/3 = 12 Ω and three of them in parallel give 12/3 = 4 Ω, i.e. 36 ÷ 3².
- practice — not a real PYQ
A uniform wire is stretched so that its length doubles, its volume remaining constant. Its resistance becomes
- (a)half
- (b)double
- (c)four times
- (d)unchanged
Answer(c) four times — doubling the length halves the area at constant volume, and R = ρL/A rises by a factor of 2 × 2.