Two conducting wires of the same material and of equal length and equal diameters are first

Given that

Inhaltsverzeichnis Show

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.

We need to find out

We need to calculate the ratio of heat produced in series and parallel combination

Solution

Let us make the following assumptions to calculate the ratio of heat produced in series and parallel combination

Let us assume the equivalent resistances of the wires if connected in series as Rs

Let us assume the equivalent resistances of the wires if connected in parallel as Rp

Formula

Total resistance R = R1 + R2 + …………..

So calculating the series resistance in the given combination we get

RS =R+R=2R————-(i)

Total resistance 1/R = 1/R1 + 1/R2 + …………..

So calculating the parallel resistance in the given combination we get

1/ RP = 1/R +1/R

1/ RP = 2/R

RP =R/2———-(ii)

To find the ratio we will combine equations (i) and (ii) we get

Rp/Rs = (R/2)/2R = 1/4

The ratio of heat produced is 1/4

Answer

Hence the ratio of the resistances is 1/4

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Text Solution

`1 : 2``2 : 1``1 : 4``4 : 1`

Solution : Since both the wires are made of the same material and have equal lengths and equal diameters, these have the same resistance. Let it be `R`. When connected in series , their equivalent resistance is given by `R_s = R + R = 2 R` When connected in parallel, their equivalent resistance is given by `(1)/(R_p) = (1)/( R) + (1)/( R) = (2)/(R)` or `R_p = (R )/(2)` Further, electrical power is given by `P = (V^2)/( R)` Power (or heat produced) in series, `P_s = (V^2)/(R_s)` Power (or heat produced) in parallel, `P_p = (V^2)/(R_p)` Thus, `(P_s)/(P_p) = (V^2//R_s)/(V^2//R_p) = (R_p)/(R_s) = (R//2)/(2 R) = (1)/(4)` or `P_s : P_p : : 1 : 4` Thus, (c) is the correct answer.

Last updated at Oct. 7, 2021 by

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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 = 𝜌 𝑙/𝐴 Since 𝜌, 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/𝑅_𝑝 = 1/𝑅_1 + 1/𝑅_2 1/𝑅_𝑝 = 1/𝑅 + 1/𝑅 1/𝑅_𝑝 = 2/𝑅 Rp = 𝑅/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 = (𝑉/𝑅)^2Rt H = 𝑉^2/𝑅^2 × Rt H = 𝑉^2/𝑅 × t We know that V = IR I = 𝑉/𝑅 Series HS = 𝑉^2/𝑅_𝑆 t HS = 𝑉^2/((2𝑅)) t HS = (𝑉^2 𝑡)/2𝑅 Parallel Hp = 𝑉^2/𝑅_𝑝 t Hp = 𝑉^2/(𝑅/2) t HS = (2 𝑉^2)/𝑅 t Finding ratio 𝐻_𝑆/𝐻_𝑝 = (((𝑉^2 𝑡)/(2 𝑅)))/(((2𝑉^2 𝑡)/𝑅) ) 𝐻_𝑆/𝐻_𝑝 = (𝑉^2 𝑡)/(2 𝑅) × 𝑅/(2 𝑉^2 𝑡) 𝐻_𝑆/𝐻_𝑝 = 1/4 ∴ HS : Hp = 1 : 4 Hence, the ratio is 1 : 4 ∴ Correct answer is (c)