contestada

at what rate, in megawatts, must energy be dissipated when the full current is switched off in 65 ms?

Respuesta :

The energy dissipation should be 65 Kilo joule.

Power is nothing but measurement the rate of electrical energy transfer by an electric circuit per unit of time. It is denoted by P and measured using the SI unit of power which is watt or one joule per second. Electric power is commonly supplied by electric batteries and produced by electric generators.

So, P = ΔE/Δt

here P = 1 megawatt and time is 65 x 10^-3 sec

putting the values we get that

1 x 10^6 =  ΔE/ 65 x 10^-3

ΔE = 10 ^ 6 x 65 x 10 ^-3

ΔE = 65 x 10 ^3

ΔE = 65 K Joule

hence the energy dissipation should be 65 Kilo joule.

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