Question

an electric iron consumes energy at a rate of 840 watt when heating is at the maximum rate and 360 what when the heating is at the minimum the voltage is 220 V. What the current and the resistance in each case ​

Answer 1

Answer:

When heating at max rate,

power, P = 840W

V = 220V

We know P = VI

⇒ 840 = 220 × I

⇒ I = 840/220

⇒ I = 4 A

R = V/I = 220/4 = 55 Ω

So current = 4A and resistance = 55Ω

When heating at minimum rate,

power, P = 360W

V = 220V

We know P = VI

⇒ 360 = 220 × I

⇒ I = 360/220

⇒ I = 1.636 A

R = V/I = 220/1.636 = 134.45 Ω

So current = 1.636A and resistance = 134.45Ω

Answer 2

ʜᴇʏᴀ ᴍᴀᴛᴇ!!

ʜᴇʀᴇ ɪs ᴜʀ ᴀɴsᴡᴇʀ ⏬ ⏬

ᴀs ᴡᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ ᴘᴏᴡᴇʀ ɪɴᴘᴜᴛ ɪs ᴘ = ᴠ ɪ

ᴛʜᴜs ᴛʜᴇ ᴄᴜʀʀᴇɴᴛ ɪ = ᴘ/ᴠ

(ᴀ) ᴡʜᴇɴ ʜᴇᴀᴛɪɴɢ ɪs ᴀᴛ ᴛʜᴇ ᴍᴀxɪᴍᴜᴍ ʀᴀᴛᴇ

ɪ = 840 ᴡ/220 ᴠ = 3.28 ᴀ;

ᴀɴᴅ ᴛʜᴇ ʀᴇsɪsᴛᴇɴᴄᴇ ᴏғ ᴛʜᴇ ᴇʟᴇᴄᴛʀɪᴄ ɪʀᴏɴ ɪs

ʀ = ᴠ/ɪ = 220 ᴠ/ 3.82 ᴀ = 57.60

(ʙ) ᴡʜᴇɴ ʜᴇᴀᴛɪɴɢ ɪs ᴀᴛ ᴍɪɴɪᴍᴜᴍ ʀᴀᴛᴇ,

ɪ = 360 ᴡ/ 220 ᴠ = 1.64 ᴀ;

ᴀɴᴅ ʀᴇsɪsᴛᴇɴᴄᴇ ᴏғ ᴛʜᴇ ᴇʟᴇᴄᴛʀɪᴄ ɪʀᴏɴ ɪs

ʀ = ᴠ/ɪ = 220 ᴠ/1.64 ᴀ = 134.15