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Question

An alternating current is given by i = i1 cos ωt + i2 sin ωt. The rms current is given by

(a) i1+i22
(b) i1+i22
(c) i12+i222
(d) i12+i222

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Solution

(c) i12+i222
Given:
i = i1 cos ωt + i2 sin ωt
The rms value of current is given by,
irms=0Ti2dt0Tdti=i1cos ωt +i2 sin ωtirms=0Ti1cos ωt +i2 sin ωt2dt0Tdtirms=0Ti12cos2ωt +i22 sin2ωt +2i1i2sin ωt cos ωt dt0Tdtirms=0Ti12(cos 2ωt+1)2 +i22 (1-cos 2ωt) 2 +i1i2sin 2ωt dt0Tdt [cos2ωt =(cos 2ωt+1)2 , sin2ωt=(1-cos 2ωt) 2 ]

We know that, T = 2π

Integrating the above expression
irms=12i12 02π1dt+02πcos 2ωt dt + i22 02π1 dt -02π cos 2ωt dt +i1i202π sin 2ωt dt02πdt

The following integrals become zero
02πcos 2ωt dt = 0=02π sin 2ωt dt
Therefore, it becomes
irms=i12202π1dt+i22202π1dt02πdtirms=i122×2π +i222×2π2πirms =i12+i222

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