Finding resistance/Power Consumption

[glasgowm]

An electric cooker has two settings, high and low. It takes 1A at the low setting and 3A at the high setting (It lists 230 as the voltage on the diagram)

Find the resistance of R1 & R2 (they are parallel circuits)

V=IR (R= V/I)
high setting: R=V/I ... R = 230V/3A = 76.6667Ω
low setting: R=V/I ... R = 230V/1A = 230Ω

I figure I'm doing something wrong since the answers I come out with a bit weird

Part 2

Calculate the power consumption

P = I²R

P = 1²A*230 | P = 3²A*76.6

 

[Doc Al]

high setting: R=V/I ... R = 230V/3A = 76.6667Ω
low setting: R=V/I ... R = 230V/1A = 230Ω

I figure I'm doing something wrong since the answers I come out with a bit weird

Looks good to me.

Part 2

Calculate the power consumption

P = I²R

P = 1²A*230 | P = 3²A*76.6

Also good. But complete the calculation and assign proper units.

(Power consumption can also be calculated using P = VI.)

 

[m2003]

P = 230W | P = 690W

Your calculations for the resistance of R1 and R2 are correct. The reason the values seem a bit strange is because they are very small resistances, which is common for household appliances. Resistance is measured in ohms (Ω) which is a unit of electrical resistance. In this case, R1 has a resistance of 76.6667Ω and R2 has a resistance of 230Ω.

For the power consumption, your calculations are also correct. The power consumed by the electric cooker is directly proportional to the resistance and the square of the current flowing through it. This is why the power consumption at the high setting is three times greater than at the low setting. This information is important for understanding the energy usage and efficiency of the electric cooker.