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
Why
Correct — A, 2 ohm. Cutting the 50 Ω wire into five equal lengths gives five pieces, each of resistance 50 Ω ÷ 5 = 10 Ω, because resistance is proportional to length. Connecting n equal resistors of resistance R in parallel gives an equivalent resistance R/n, so five 10 Ω resistors in parallel give 10 Ω ÷ 5 = 2 Ω.
Why the others are wrong
- (b)10 ohm — 10 Ω is the resistance of just one of the five pieces; the question asks for all five connected in parallel, which is 10/5 = 2 Ω.
- (c)0·5 ohm — This would require dividing 10 Ω by 20 (or some similar slip); five equal 10 Ω resistors in parallel give 10/5 = 2 Ω.
- (d)5 ohm — 5 Ω comes from dividing the original 50 Ω by 10, or from mis-counting the pieces; each piece is 10 Ω and five in parallel give 2 Ω.
Concept
The resistance of a uniform wire is proportional to its length, so cutting a wire into n equal parts divides its resistance by n. For n equal resistors of resistance R connected in parallel, the equivalent resistance is R/n, always smaller than a single resistor.
This is a two-step problem. First find the resistance of each cut piece (50/5 = 10 Ω), then combine the five pieces in parallel (10/5 = 2 Ω). The common error is stopping after the first step.
Key facts
- Resistance is directly proportional to length, so cutting a wire into n equal parts divides its resistance by n.
- Each of the five pieces here has 50/5 = 10 Ω.
- For n equal resistors R in parallel, the equivalent resistance is R/n.
- Five 10 Ω resistors in parallel give 10/5 = 2 Ω — less than any single resistor.
Each piece is 10 Ω; five in parallel give 2 Ω — option (a).
Study next
Common traps
- Stopping at 10 Ω (the resistance of one piece) instead of combining all five in parallel.
- Thinking parallel resistance adds up — the parallel value is always less than the smallest resistor.
Two steps in one — first find each cut piece's resistance, then combine them in parallel using R/n.
Related PYQs
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
Near-identical twin — cut a wire into equal parts, then join in parallel. That NDA item cuts 20 Ω into two (each 10 Ω, parallel → 5 Ω); this one cuts 50 Ω into five (each 10 Ω, parallel → 2 Ω).
Three equal resistors are connected in parallel configuration in a closed electrical circuit. Then the total resistance in the circuit becomes
- (a) one-third of the individual resistance.
- (b) two-third of the individual resistance.
- (c) equal to the individual resistance.
- (d) three times of the individual resistance.
Answer(a) one-third of the individual resistance.
Same rule — n equal resistors in parallel give R/n. That NDA item states three equal resistors in parallel give one-third the resistance; this one applies R/n to five equal cut pieces.
Practice
- practice — not a real PYQ
A wire of resistance 24 Ω is cut into three equal parts and the three parts are joined in parallel. The equivalent resistance is
- (a)24 Ω
- (b)8 Ω
- (c)2·67 Ω
- (d)72 Ω
Answer(c) 2·67 Ω — each part is 8 Ω, and three in parallel give 8/3 ≈ 2·67 Ω. - practice — not a real PYQ
Two equal resistors of 6 Ω each are connected in parallel. Their equivalent resistance is
- (a)12 Ω
- (b)6 Ω
- (c)3 Ω
- (d)2 Ω
Answer(c) 3 Ω — R/n = 6/2.