Question 683: Current Electricity

Question

An electric kettle has two heating coils. When one of the coils is connected to an a.c. source, the water in the kettle boils in 10 minutes. When the other coil is used the water boils in 40 minutes. If both the coils are connected in parallel, the time taken by the same quantity of water to boil will be

Accepted Answer

Let R
1
and R
2
be the resistance of the two coils and V be the voltage supplied.
Effective resistance of two coils in parallel = $${{{R_1}{R_2}} \over {{R_1} + {R_2}}}$$
Let H be the heat required to begin boiling in kettle.
Then H = Power $$ 	imes $$ time = $${{{V^2}{t_1}} \over {{R_1}}} = {{{V^2}{t_2}} \over {{R_2}}}$$
For parallel combination, H = $${{{V^2}\left( {{R_1} + {R_2}} ight){t_p}} \over {{R_1}{R_2}}}$$
$$ \Rightarrow {1 \over {{t_p}}} = \left( {{{{t_2} + {t_1}} \over {{t_2}{t_1}}}} ight)$$
$$ 	herefore {t_p} = {{{t_2}{t_1}} \over {{t_2} + {t_1}}} = {{10 	imes 40} \over {10 + 40}} = 8$$ minute

Answer

8 minutes

Answer

4 minutes

Answer

25 minutes

Answer

15 minutes

Current Electricity

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