Kerala Engineering Entrance 2007 Questions on Heating Effect of Electric Current

Two questions on heating effect of electric current appeared in KEAM (Engineering) 2007 question paper:

(1) The resistance of a wire at room temperature 30º C is found to be 10 Ω. Now to increase the resistance by 10%, the temperature of the wire must be [Temperature coefficient of resistance of the material of the wire is 0.002 per º C]

(a) 36º C (b) ) 83º C (c) ) 63º C (d) ) 33º C (e) ) 66º C

The resistance (R_t) at tº C can be expressed in terms of the resistance (R₀) at 0º C and the temperature coefficient (α) of resistance as

R_t = R₀(1 + αt).

Therefore, we have

R₃₀ = 10 = R₀(1 + 0.002×30) and

R_t = 11 = R₀(1 + 0.002 t)

[The resistance R_t at the unknown temperature ‘t’ is greater by 10% and is therefore equal to (10 + 1) Ω = 11 Ω]

From the above equations we have

10/11 = (1 + 0.06)/ (1 + 0.002t) from which t = 83º C.

(2) If R₁ and R₂ be the resistances of the filaments of 200 W and 100 W electric bulbs operating at 220 V, then R₁/R₂ is

(a) 1 (b) 2 (c) 0.5 (d) 4 (e) 0.25

Since the power is V²/R where V is the operating voltage and R is the resistance, we have

220²/R₁ = 200 and

220²/R₂ = 100.

Dividing the second equation by the first, we obtain

R₁/R₂ = 0.5