QJEEM24Phb69d: JEE Main 2024 Physics - Class 12 JEEM24Phb69d Solution: A alternating current at any instant is given by A. The value…

The given alternating current (AC) can be represented as $i=6+\sqrt{56} \sin (100 \pi t+\pi / 3)$ A, where $6$ is the DC component and $\sqrt{56} \sin (100 \pi t+\pi / 3)$ is the AC component of the current. The RMS (Root Mean Square) value of an alternating current is a measure of the equivalent direct current (DC) that will produce the same power in a resistor. The RMS value is mostly relevant for the AC component of the current, as the DC component's effective value is just its magnitude itself.

The RMS value of the total current is not straightforward because the presence of the DC component affects how we calculate the RMS value. However, when calculating RMS values for a signal consisting of a superposition of AC and DC components, one notable property is that the RMS value of the combined signal is the square root of the sum of the squares of the RMS values of the separate AC and DC components.

First, let's acknowledge the components separately:

• The DC component is: $6$ A


• The AC component is: $\sqrt{56} \sin (100 \pi t+\pi / 3)$ A

For the DC component, the RMS value is simply its magnitude:

$I_{RMS, DC} = 6$ A

For the AC component, the RMS value is calculated using the formula for the RMS value of a sinusoidal function, which is $I_{RMS} = \frac{I_{max}}{\sqrt{2}}$ , where $I_{max}$ is the peak value of the current. In this case, $I_{max} = \sqrt{56}$ .

Therefore, the RMS value of the AC component is:

$I_{RMS, AC} = \frac{\sqrt{56}}{\sqrt{2}} = \frac{\sqrt{56}}{\sqrt{2}} = \sqrt{\frac{56}{2}} = \sqrt{28}$ A.

Finally, to find the total RMS value of the current, combine the DC and AC components as follows:

$I_{RMS} = \sqrt{{(I_{RMS, DC})}^2 + {(I_{RMS, AC})}^2}$

Substituting the values:

$I_{RMS} = \sqrt{{(6)}^2 + {(\sqrt{28})}^2}$

$= \sqrt{36 + 28}$

$= \sqrt{64}$

$= 8$ A.

Therefore, the RMS value of the current is $8$ A.