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GATE 2015 EE - SET 1 - Question 45

Solution :

I_{dc}= 8/1 = 8A

\left ( I_{AC} \right )_{P} =  \frac{12}{\sqrt{2}} = 8.485A

I_{rms}= \sqrt{I_{dc}^{2}} + \left [ \frac{I_{AC}}{\sqrt{2}} \right ]^{2}

Substituting all the values in

I_{rms} = 9.99A = 10 A

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