Answer to Question #77733 in Electric Circuits for Mohannad

Question

An electric kettle is rated at 2kW when connected to a 230V electrical supply.
a) Calculate the current that will flow when the kettle is turned on.
b) What fuse should be included in the circuit of the kettle? Assume that the fuses available are 3A, 5A,
and 13A.
c) Calculate the resistance of the heating element of the kettle.
d) If 420kJ of energy are required to boil some amount of water, how much time would the kettle need
to boil this amount of water?

Answer 1

a) We can find the current that will flow when the kettle is turned on from the definition of the electrical power:
P=IV,
here, I is the current that will flow when the kettle is turned on, V is the electrical supply voltage.
Then, we get:
I=P/V=(2∙〖10〗^3 W)/(230 V)=8.7 A.
b) Since, the current that will flow when the kettle is turned on is 8.7 A we need to include 13 A fuse in the circuit of the kettle.
c) We can find the resistance of the heating element of the kettle from the Ohm’s law:
R=V/I=(230 V)/(8.7 A)=26.4 Ω.
d) We can find the time needed to boil some amount of water in the kettle from the formula:
Q=Pt,
here, Q is the amount of energy required to boil some amount of water in the kettle, P is the electrical power of the kettle and t is the time needed to boil some amount of water in the kettle.
Then, we get:
t=Q/P=(4.2∙〖10〗^5 J)/(2∙〖10〗^3 W)=210 s.