An electric kettle has a coiled wire inside that dissipates power when it carrie

Question

An electric kettle has a coiled wire inside that dissipates power when it carries a current, warming the water in the kettle. A kettle designed for use in England carries 13 A when connected to a 230 V outlet. A. What is the resistance of the wire? (Use Ohm’s law) B. What power is dissipated when the kettle is running? (You have I,V and R to find P) The kettle is now redesigned to work at the lower voltage; Now, the kettle carries 13 A when connected to 120 V. C. What is the new resistance of the wire? (use Ohm’s law)

Explanation / Answer

Given,

I = 13 A ; V = 230 V

A) we need to find the resistance "R" of the wire.

We know from Ohm's law that,

V = I R=> R = V/I

R = 230/13 = 17.7 Ohm

Hence, R = 17.7 Ohm.

B)We know that the power dissipated is given by:

P = I2R = (13)2 x 17.7 = 2991.3 Watts

Hence, P = 2991.3 Watts.

C) I' = 13 A and V' = 120 V ; R' = ?

again from Ohm's law

V = I R => R = V/I

R' = V'/I' = 120/13 = 9.23 Ohm

Hence, R' = 9.23 Ohm.

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