An electrical appliance of power 2KW …

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Yogita Ingle 5 years, 8 months ago
The power of the electrical appliance = 2kw (when changed to watts it shall be 2000w)
The potential difference= 220v
Since, I=P/V, we will observe
2000/220= 9.09v
Thus, the electrical appliance requires a fuse of more than 5A else, with 9.09v the electric fuse will melt and the circuit will get broken, as it is drawing more current than its safe limit.
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