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