Solution:
The two currents can be vectorially added with phase difference \(\phi = 2\pi/3 = 120^\circ\). Resultant amplitude is \(I_0 = \sqrt{4^2 + 4^2 + 2 \times 4 \times 4 \cos 120^\circ} = 4\).
The peak value of the alternating current given by \(I = 4 \sin\omega t + 4 \sin(\omega t + 2\pi/3)\) is:
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