Question 2
An electric heater is to be made from nichrome wire.
Nichrome has a resistivity of
1.0 × 10-6 Ω m at the operating temperature of the
heater.
The heater is to have a power dissipation of 60 W when
the potential difference across its terminals is 12 V.
(a) For the heater operating at its designed power,
(i) calculate the current, [2]
(ii) show that the resistance of the
nichrome wire is 2.4 Ω. [2]
(b) Calculate the length of nichrome wire of diameter 0.80 mm
required for the heater. [3]
(c) A second heater, also designed to operate from a 12 V
supply, is constructed using the same nichrome wire but using half the length
of that calculated in (b).
Explain quantitatively the effect of this change in
length of wire on the power of the
heater. [3]
Reference: Past Exam Paper – June 2010 Paper 21 Q6
Solution:
(a)
(i)
Power P = VI
60 = 12 × I
Current
I (= 60 / 12) = 5(.0) A
(ii)
{We can use different
formula to find the resistance. However, it is better to use the formula P = V2
/ R as it involves values of P and V which are already given in the question.}
EITHER V = IR or P = I2 × R or P = V2 / R
EITHER 12 = 5×R or 60 = 52 × R or 60 = 122 / R
Resistance
of nichrome wire, R = 2.4 Ω
(b)
Resistance of wire: R = ρL / A
Cross-sectional area, A (=
πr2) = π × (0.4×10-3)2 [= 5.03×10-7]
{From R = ρL / A L = RA / ρ}
Length, L = 2.4 × (5.03×10-7) / (1.0×10-6) = 1.2 m
(c)
{Since R is proportional
to L: (R = ρL / A)}
The resistance is halved
when half the length of the wire is taken.
The current is doubled
{since current is inversely proportional to the resistance (I = V / R)}
{P = VI. Since the current is doubled, the power P is also doubled. V is NOT halved. Instead V = 12 V. (sum of e.m.f. = sum of p.d. in circuit)}
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