Physics, asked by shameerkhanpallikala, 6 months ago

power of a 200V electrical device is 500W.
calculate the electric current. through this device.
what will amparage of the fuse wire suitable for this device.​

Answers

Answered by Anonymous
13

Answer:

Fuses are safety devices that are to be built into our electrical system. If there were no fuses and we operated too many appliances on a single circuit, the cable carrying the power for that circuit would get extremely hot, short circuit, and possibly start a fire. To prevent electrical overloads, fuses are designed to trip or blow, stopping the flow of current to the overloaded cable.

The fuse must always be connected to the mains and it must be of correct value. For example, a 15-ampere fuse should trip when the current through it exceeds 15 amperes. A 20-ampere fuse should blow when the current through it exceeds 20 amps.

In this case, let us consider a device of power 5 kW, that is, 5000 W and 200 V.

The current through the appliance is given as I=P/V.

That is, I=

V

P

=

200

5000

=25A.

The fuse of 8 A cannot be used because the current through the appliance is

greater the fuse rating. When the 25 A current flows through the fuse it will

blow off and it will not be able to fulfill it purpose.

Hence, a fuse is rated 8 A cannot be used with this appliances.

stay blessed ✨

Answered by Qwdelhi
0

The electric current through the given device is 2.5 A.

Given:

V = 200 V, P = 500W.

To Find:

The electric current.

Solution:

We know that

I = P / V

= 500 / 200

= 5/2

= 2.5 A.

Therefore, The electric current through the given device is 2.5 A.

Amperage is the ratio of the power of equipment to the voltage applied., which is equal to the current.

When we apply a high current to the circuit, it may damage the circuit. To avoid this, we should maintain a current less than equal to the amperage of the fuse wire.

Hence, The suitable amperage of the fuse wire will be 2.5 A.

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