For an electric iron consuming 1 kW of electric power at 220 V, the current flowing in the circuit is (1000/220) A, which is approximately 4.54 A. In this case, a 5 A fuse must be used to ensure that the fuse can handle the current without breaking the circuit during normal operation.
Provide an example of how the selection of a fuse rating is determined based on the power consumption of a device, such as an electric iron.
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For example, consider an electric iron with a power rating of 1200 watts and an operating voltage of 120 volts. Using the formula P = VI (power equals voltage multiplied by current), the current drawn by the iron is P/V, which is 1200/120 = 10 amperes. To ensure proper protection, a fuse rating slightly higher than the device’s current is chosen. In this case, a 15-ampere fuse may be selected, providing a safety margin. This way, if the iron experiences a temporary surge or malfunction, the fuse will blow, disconnecting the circuit and preventing damage to the iron or potential fire hazards.