An electrical iron uses a power of 1150 W when set to higher temperature. If set to lower temperature, it uses 460 W power. Find out the electric current and the respective resistances for the two settings, if the electric iron is connceted to a potential difference of 230 V.
Step 1 : The problem deals with two cases :
i. Electric iron at higher temperature
ii. Electric iron at lower temperature
Step 2 : Key factors to understand :
i. In the above two cases, Power, electric current and the resistances of the electric iron will be different due to their different setting in temperature.
ii. Potential difference drawn by the electric iron will be same in both the cases,

Explanation

Step 3 : At higher temperature,
Power consumed `(P_(1))=1150 W,` Potential difference (V) = 230 V
Current `(I_(1))` = ?, Resistance of the electric iron `(R_(1))=?`
For current , `P_(1)=Vxx I_(1)`
`therefore" "I_(1)=(P_(1))/(V)=(1150)/(230)=5A`
For resistance, `V=I_(1)R_(1)`
`therefore " "R_(1)=(V)/(I_(i))=(230)/(5)=46Omega`
Hence, At higher temperature, the electric current flowing through it is 5A and the resistance of the electric iron of `46Omega`.
Step 4 : At lower temperature,
Power consumed `(P_(2))=460W,` Potential difference `(V)=230V`
Current `(I_(2))=?`, Resistance of the electric iron `(R_(2))=?`
For current, `P_(2)=VxxI_(2)`
`therefore" "I_(2)=(P_(2))/(V)=(460)/(230)=2A`
For resistance , `V=I_(2)R_(2)`
`therefore" "F_(2)=(V)/(I_(2))=(230)/(2)=115Omega`
Hence, At lower temperature, the electric is flowing through it 2 A and the resistance of the electric iron is `115Omega`.