A direct current of 10 ,A is superimposed on | vedclass

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(D) The total current is given by $I = I_{DC} + I_{AC} = 10 + 40 \cos \omega t$.The effective $(RMS)$ value of a current $I = I_{DC} + I_0 \cos \omega

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$30$

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(D) The total current is given by $I = I_{DC} + I_{AC} = 10 + 40 \cos \omega t$.The effective $(RMS)$ value of a current $I = I_{DC} + I_0 \cos \omega t$ is given by the formula $I_{rms} = \sqrt{I_{DC}^2 + I_{rms, AC}^2}$.Here,$I_{DC} = 10 \, A$ and the peak value of the $AC$ component is $I_0 = 40 \, A$.The $RMS$ value of the $AC$ component is $I_{rms, AC} = \frac{I_0}{\sqrt{2}} = \frac{40}{\sqrt{2}} = 20\sqrt{2} \, A$.Now,calculate the effective value of the resulting current:$I_{rms} = \sqrt{(10)^2 + (20\sqrt{2})^2}$$I_{rms} = \sqrt{100 + (400 	imes 2)}$$I_{rms} = \sqrt{100 + 800}$$I_{rms} = \sqrt{900} = 30 \, A$.

$A$ direct current of $10 \,A$ is superimposed on an alternating current $I = 40 \cos \omega t \; (A)$ flowing through a wire. The effective value of the resulting current will be ....... $A$.

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