Question

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

Answer 1

Answer:

Refer to the attachment

Answer 2

Answer:

`"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

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