Question

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Answer 1

Answer:

An electric iron consumes energy at a rate of 840 W when heating is at the maximum rate and 360 W when the heating is at minimum. The voltage is 220 V.

Answer 2

Question :

An electric iron consumes energy at a rate of 840W when heating is at maximum rate and 360W when heating is at the minimum. The voltage is 220V. What are the current and the resistance in each case?

Solution :

Given :

• Voltage [V] = 220V
• Max. power = 840W
• Min. power = 360W

C A S E ( A ) :

Here :

• Power [P] = 840W
• Voltage [V] = 220V

Finding current :

• P = VI
• I = P/V
• I = 840/220
• I = 84/22
• I = 3.82A

Finding resistance :

• V = IR
• R = V/I
• R = 220/3.82
• R = 57.60Ω

The current and resistance is 3.82A and 57.6Ω.

C A S E ( B ) :

Here :

• Power [P] = 360W
• Voltage [V] = 220V

Finding current :

• I = P/V
• I = 360/220
• I = 36/22
• I = 18/11
• I = 1.64A

Finding resistance :

• R = V/I
• R = 220/1.64
• R = 134.15Ω

Therefore, the current and resistance is 1.64A and 134.15Ω.

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Hope this helps! :)