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A. 6 ampere

B. 2 ampere

C. $2 \sqrt{3}$ ampere

D. 0.65 ampere

Answer

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$H=I^2 Rt$

Alternating current has fluctuating current amplitudes. If we have a DC supply with a value I, it will constantly provide us current of magnitude I, but if we have an alternating current, then we will obtain the amplitude as $I_0 \sin \omega t$. Thus, we get a sinusoidal supply if we use alternating current. We know, sine has an average of zero over one complete cycle. Therefore, a parameter called rms or root mean square value of current is used. This rms value is what the ammeters are able to measure. It (rms) is responsible for heating a device or a wire when alternating current is passed through it.

Therefore, when passing alternating current through a resistor R for time t, we may write:

$H= I_{rms}^2 Rt$

We know, the DC version of this is just $H=I^2 Rt$. But, we are given that Heat produced is thrice of the amount that a 2A DC source produces. So, first we get the DC source heat as:

$H_{dc} =2^2 Rt$

$H_{dc} =4Rt$

Then, thrice of this is just

H=12Rt.

Equating this heat with the one produced by ac current:

$I_{rms}^2 Rt= 12Rt $

Therefore,

$I_{rms} = \sqrt{12} A$

$I_{rms} = 2 \sqrt{3}A$