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Hint
The rms value of the current of an alternating current is responsible for producing heat. The formula for heat produced in resistor R, when I current is passed it from time it is $H = {I^2}Rt$.
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 the value I, but if we had a alternating current then we get a sinusoidal supply and sine has an average value of zero over a cycle therefore we use rms parameter is used.it is responsible for heating the device when alternating current passed through it.
Therefore, when passing alternating current through resistor R for time t heat will be equal to
${H_1} = I_{rms}^2Rt$
And we also know that for DC current, heat can be written as ${H_2} = {I^2}Rt = {2^2}Rt = 4Rt$
As it is given that the heat produced by the alternating current is three times to the heat produced by DC current. I.e. ${H_1} = 3{H_2}$
On substituting the values, we get
$ \Rightarrow I_{rms}^2Rt = 3 \times 4Rt$
$ \Rightarrow I_{rms}^2Rt = 12Rt$
$ \Rightarrow I_{rms}^2 = \sqrt {12} = 2\sqrt 3 A$
Hence, the rms value of current is $2\sqrt 3 A$
Therefore, option (D) is correct.
Additional Information
There is a quantity called peak value of an alternating current defined as $I_0$.
This is the maximum value of current that it can achieve. It is related to rms as ${I_{rms}} = \dfrac{{{I_0}}}{{\sqrt 2 }}$.
Note
one must be careful with thrice part of the question. Thus, when we equate the thrice part of DC heat to AC heat, we must not mix up with that with DC heat.
The rms value of the current of an alternating current is responsible for producing heat. The formula for heat produced in resistor R, when I current is passed it from time it is $H = {I^2}Rt$.
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 the value I, but if we had a alternating current then we get a sinusoidal supply and sine has an average value of zero over a cycle therefore we use rms parameter is used.it is responsible for heating the device when alternating current passed through it.
Therefore, when passing alternating current through resistor R for time t heat will be equal to
${H_1} = I_{rms}^2Rt$
And we also know that for DC current, heat can be written as ${H_2} = {I^2}Rt = {2^2}Rt = 4Rt$
As it is given that the heat produced by the alternating current is three times to the heat produced by DC current. I.e. ${H_1} = 3{H_2}$
On substituting the values, we get
$ \Rightarrow I_{rms}^2Rt = 3 \times 4Rt$
$ \Rightarrow I_{rms}^2Rt = 12Rt$
$ \Rightarrow I_{rms}^2 = \sqrt {12} = 2\sqrt 3 A$
Hence, the rms value of current is $2\sqrt 3 A$
Therefore, option (D) is correct.
Additional Information
There is a quantity called peak value of an alternating current defined as $I_0$.
This is the maximum value of current that it can achieve. It is related to rms as ${I_{rms}} = \dfrac{{{I_0}}}{{\sqrt 2 }}$.
Note
one must be careful with thrice part of the question. Thus, when we equate the thrice part of DC heat to AC heat, we must not mix up with that with DC heat.
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