Question

what is the approximate peak value of an alternating current producing four times the heat produced per second by a Steady Current of 2 ampere in register

Answer 1

hhDC power is given as:

Pdc=I2∗R

So the power dissipated in this case, where the current is 2A is:

Pdc=4∗R

Now, as heat is energy, it will be given by power multiplied by time for which the circuit runs. In this case, I am assuming that the both the circuits run for the same amount of time and it is much larger than the time period of the AC signal applied.

So, power dissipation of an AC circuit with a resistor is given as:

Pac=I2rms∗R

Where Irms

is the root mean square value of the AC current.

With the assumptions made above, as we want four times the heat produced in DC circuit we need:

Pac=4∗Pdc

I2rms=4∗4

Irms=4A

If your AC signal is sinusoidal,

Ip=1.414∗4=5.656A