A piece of wire of resistance R is cut into five 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) 1/25 (b) 1/5 (c) 5 (d) 25

Transcript

NCERT Question 1 A piece of wire of resistance R is cut into five 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) 1/25 (b) 1/5 (c) 5 (d) 25
We know that,
R ∝ l
Initial resistance of the wire = R Ω
When wire is cut into 5 pieces, its resistance will become 1/5 th
Resistance of each piece of wire after cut into 5 pieces = 𝑅/5 Ω
Now, these wires are connected in parallel
We know that,
in parallel connection,
equivalent resistance is given by
1/𝑅′ = 1/𝑅_1 + 1/𝑅_2 + 1/𝑅_3 + 1/𝑅_4 + 1/𝑅_5
1/𝑅′ = 1/(𝑅/5) + 1/(𝑅/5) + 1/(𝑅/5) + 1/(𝑅/5) + 1/(𝑅/5)
1/𝑅′ = 5/𝑅 + 5/𝑅 + 5/𝑅 + 5/𝑅 + 5/𝑅
1/𝑅′ = 25/𝑅
R’ = 𝑅/25
Finding 𝑹/𝑹′
𝑅/𝑅′ = 𝑅/(𝑅/25)
𝑅/𝑅′ = (25 𝑅)/𝑅
𝑅/𝑅′ = 25
Answer is (d) 25

CA Maninder Singh is a Chartered Accountant for the past 12 years and a teacher from the past 16 years. He teaches Science, Economics, Accounting and English at Teachoo

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