Answer
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Hint: The rms value or the root mean square value of an alternating current is responsible for producing heat. If I have a resistor, then it will get heated according to the rms value that the alternating current supplied to it will hold.
Formula used: The formula for heat produced in a resistor R, when I current is passed through it for time t is given as:
$H=I^2 Rt$
Complete step by step answer:
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,
$I_{rms} = \sqrt{12} A$
$I_{rms} = 2 \sqrt{3}A$
So, the correct answer is “Option C”.
Additional Information: There is a quantity called peak value of an alternating current defined as $I_0$. This is the maximum value that the ac current can achieve. It is related to rms value as $I_{rms} = \dfrac{I_0}{\sqrt{2}}$. In our domestic supply of 220V, 5A, we have 5A as rms value of the ac supply current. Peak value of such supply is about 7.07A.
Note: One must be careful with the thrice part of the question. One can also use primed variables for Heat expression to denote DC heat of 2A current separately. Thus, when we equate the thrice of DC heat to AC heat, we don't mix up that with the DC heat.
Formula used: The formula for heat produced in a resistor R, when I current is passed through it for time t is given as:
$H=I^2 Rt$
Complete step by step answer:
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,
$I_{rms} = \sqrt{12} A$
$I_{rms} = 2 \sqrt{3}A$
So, the correct answer is “Option C”.
Additional Information: There is a quantity called peak value of an alternating current defined as $I_0$. This is the maximum value that the ac current can achieve. It is related to rms value as $I_{rms} = \dfrac{I_0}{\sqrt{2}}$. In our domestic supply of 220V, 5A, we have 5A as rms value of the ac supply current. Peak value of such supply is about 7.07A.
Note: One must be careful with the thrice part of the question. One can also use primed variables for Heat expression to denote DC heat of 2A current separately. Thus, when we equate the thrice of DC heat to AC heat, we don't mix up that with the DC heat.
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