An electrical iron uses a power of 1100 W when set to higher temperature. If set to lower temperature, it uses 330 W power. Find out the electric current and the respective resistances for the two settings. The iron is connected to a potential difference of 220 V.

Explanation

Given: potential difference = 220 V.
Power P = (A) 1100 W, (B) = 330 W.
A. Power = 1100 W.
`I _(1) = (P)/(v) = (1100)/(220) = 5A `
B. Power = 300W
`I_(2) = (P)/(V) = (330)/(220) = 1.5A`
Resistance `R^(1) = (V)/(I_(2)) = (220)/(5) = 44Omega `
Resistance `R_(2) = (V)/(I_(2)) = (220)/(1.5)=146Omega`