Saturday, September 22, 2007

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

Rt = R0(1 + αt).

Therefore, we have

R30 = 10 = R0(1 + 0.002×30) and

Rt = 11 = R0(1 + 0.002 t)

[The resistance Rt 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

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