The currents i1 and i2 in A.C circuit are given as (in amp) : i1 =4sin(ωt−π3) and i2=4sin(ωt+π3). The r.m.s value of i3 is:

i3=i1+i2, by phasor methodi3=4 sin ωtir.m.s=42=22

i3=i1+i2, by phasor methodi3=4 sin ωtir.m.s=42=22

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The currents $i_{1} a n d i_{2}$ in A.C circuit are given as (in amp) : $i_{1} = 4 \sin (ω t - \frac{π}{3}) a n d i_{2} = 4 \sin (ω t + \frac{π}{3}) .$ The r.m.s value of $i_{3}$ is:

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$2 \sqrt{6}$

$2 \sqrt{2}$

$\sqrt{6}$

$3 \sqrt{2}$

answer is D.

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$i_{3} = i_{1} + i_{2}, b y p h a s o r m e t h o d$

$i_{3} = 4 \sin ω t$

$i_{r . m . s} = \frac{4}{\sqrt{2}} = 2 \sqrt{2}$

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