Question 401: Current Electricity

Question

The resistance of an electrical toaster has a temperature dependence given by R(T) = R
0
[1 + $$\alpha $$(T − T
0
)] in its range of operation. At T
0
= 300 K, R = 100 $$\Omega $$ and at T = 500 K, R = 120 $$\Omega $$. The toaster is connected to a voltage source at 200 V and its temperature is raised at a constant rate
from 300 to 500 K in 30 s. The total work done in raising the temperature is :

Accepted Answer

Given,
R = R
6
[1 + $$\alpha $$ (T $$-$$ t
0
)]
120 = 100 [1 + $$\alpha $$ (500 $$-$$ 300)]
$$ \Rightarrow $$   200 $$\alpha $$ = $${1 \over 5}$$
$$ \Rightarrow $$   $$\alpha $$ = 10
$$-$$3
o
C
$$-$$1
Temperature of the toaster raised from  300 K to 500 K in 30 s.
$$ 	herefore $$   Increment in the temperature in time t,
$$\Delta $$T = $${{500 - 300} \over {30}}t$$
= $${{200} \over 3}t$$
= $${{20} \over 3}t$$
Total work done in raising the temperature
= $$\int\limits_0^t {{{{V^2}} \over {R\left( t ight)}}\,dt} $$
= $$\int\limits_0^t {{{{V^2}} \over {{R_0}\left( {1 + \alpha \Delta t} ight)}}} \,dt$$
= $$\int\limits_0^{30} {{{{{\left( {200} ight)}^2}} \over {100\left( {1 + {{10}^{ - 3}} 	imes {{20} \over 3}t} ight)}}} \,dt$$
= $${{40000} \over {100}}\int\limits_0^{30} {{{dt} \over {\left( {1 + {t \over {150}}} ight)}}} $$
$$400 	imes 150\left[ {\ln \left( {1 + {t \over {150}}} ight)} ight]_0^{30}$$
$$ = 60000\left[ {\ln \left( {1 + {{30} \over {150}}} ight) - \ln 1} ight]$$
$$ = 60000\ln \left( {{6 \over 5}} ight)\,J$$

Answer

400 $$\ln \,{{1.5} \over {1.3}}\,J$$

Answer

200 $$\ln \,{{2} \over {3}}\,J$$

Answer

60000 $$\ln \,{{6} \over {5}}\,J$$

Answer

300 J

Current Electricity

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