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.

Answer

i. Electric current at higher temperature is 5 A.
ii. Electric current at lower temperature is 1.5 A.
iii. Resistance at higher temperature is `44Omega`.
iv. Resistance at lower temperature is `146.67Omega.`

Explanation

`"Given Potential difference (V) = 220 V."`
Power used at higher temperature
`(P_(1))=1100W,`
Power used at lower temperature
`(P_(1))=1100W,`
Power used at lower temperature
`(P_(2))=330W`.
`"To find: i. Electric current "(I_(1))" at higher temperature."`
`"ii Electric current "(I_(2))" at lower temperature"`
`"iii. Resistance "(R_(1))" at higher temperature"`
`"iv. Resistance "(R_(2))" at lower temperature"`
`"Formulae: i. P = VI. ii. V = 1 R"`
Calculation : For higher temperature,
From formula (i),
`I_(1)=(P_(1))/(V)=(1100)/(220)=5A`
From formula (ii),
`R_(1)=(V)/(I_(1))=(220)/(5)=44Omega`
For lower temperature,
From formula (i),
`I_(2)=(P_(2))/(V)=(330)/(220)=(3)/(2)=1.5A`
From formula (ii),
`R_(2)=(V)/(I_(2))=(220)/(1.5)=(220)/((3//2))`
`=(220xx2)/(3)=(440)/(3)=146.67Omega`