Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be –

(a) 1:2 (b) 2:1 (c) 1:4 (d) 4:1

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  1. Class 10
  2. Chapter 12 Class 10 - Electricity

Transcript

NCERT Question 4 Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be - (a) 1 : 2 (b) 2 : 1 (c) 1 : 4 (d) 4 : 1 Explanation: We know that, R = ๐œŒ ๐‘™/๐ด Since ๐œŒ, l and A are same, the resistance of two wires will be same. Let the resistance = R Series We know that, In series combination equivalent resistance is given by RS = R1 + R2 = R + R = 2 R Parallel We know that, In parallel combination equivalent resistance is given by 1/๐‘…_๐‘ = 1/๐‘…_1 + 1/๐‘…_2 1/๐‘…_๐‘ = 1/๐‘… + 1/๐‘… 1/๐‘…_๐‘ = 2/๐‘… Rp = ๐‘…/2 The potential difference in both cases in same . Let potential difference = V Time taken in both cases would be same = t We know that, Heat generated is given by H = I2Rt H = (๐‘‰/๐‘…)^2Rt H = ๐‘‰^2/๐‘…^2 ร— Rt H = ๐‘‰^2/๐‘… ร— t We know that V = IR I = ๐‘‰/๐‘… Series HS = ๐‘‰^2/๐‘…_๐‘† t HS = ๐‘‰^2/((2๐‘…)) t HS = (๐‘‰^2 ๐‘ก)/2๐‘… Parallel Hp = ๐‘‰^2/๐‘…_๐‘ t Hp = ๐‘‰^2/(๐‘…/2) t HS = (2 ๐‘‰^2)/๐‘… t Finding ratio ๐ป_๐‘†/๐ป_๐‘ = (((๐‘‰^2 ๐‘ก)/(2 ๐‘…)))/(((2๐‘‰^2 ๐‘ก)/๐‘…) ) ๐ป_๐‘†/๐ป_๐‘ = (๐‘‰^2 ๐‘ก)/(2 ๐‘…) ร— ๐‘…/(2 ๐‘‰^2 ๐‘ก) ๐ป_๐‘†/๐ป_๐‘ = 1/4 โˆด HS : Hp = 1 : 4 Hence, the ratio is 1 : 4 โˆด Correct answer is (c)

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