Question

An electric kettle draws a current of 5.2A when connected to a 240V source. What is (a) the resistance

of the heating element of the kettle. (b) The power rating of the kettle? (C) How much energy is used in
30 minutes?
[1248 W, 2.25 x106 J]
Show your work:​

Answer 1

A. 46.15 ohms

B. 1248 W

C. 2.25×10⁶ J

Step-by-step explanation:

A. Determination of the resistance.

Current (I) = 5.2 A

Voltage (V) = 240 V

Resistance (R) =?

From ohm's law,

V = IR

240 = 5.2 × R

Divide both side by 5.2

R = 240 / 5.2

R = 46.15 ohms

B. Determination of the power.

Current (I) = 5.2 A

Voltage (V) = 240 V

Power (P) =?

P = IV

P = 240 × 5.2

P = 1248 W

C. Determination of the energy.

Power (P) = 1248 W

Time (t) = 30 minutes

Energy (E) =?

Next, we shall convert 30 minutes to seconds. This can be obtained as follow:

1 min = 60 s

Therefore,

30 min = 30 min × 60 s / 1 min

30 min = 1800 s

Finally, we shall determine the energy. This can be obtained as illustrated below:

Power (P) = 1248 W

Time (t) = 1800 s

Energy (E) =?

E = Pt

E = 1248 × 1800

E = 2.25×10⁶ J