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Compute the RMS (root-mean-square) value of the current i(t) = 10 [1 + sin(-t)] A over one full cycle.

Difficulty: Medium

Correct Answer: √150 A (≈ 12.25 A)

Explanation:


Introduction / Context:
RMS value measures the effective heating (power) equivalence of a time-varying current. For a waveform that consists of a DC component plus a sinusoid, the RMS can be found by combining the RMS of each orthogonal component in quadrature.


Given Data / Assumptions:

  • i(t) = 10[1 + sin(-t)] A.
  • Sinusoid is steady and periodic; evaluate over one full cycle.
  • sin(-t) = -sin t (odd symmetry), so i(t) = 10 − 10 sin t.


Concept / Approach:
For i(t) = I_DC + I_m sin(ωt + φ), the RMS value satisfies I_RMS^2 = I_DC^2 + (I_m/√2)^2 because the DC and sinusoidal AC components are orthogonal (zero cross term over a full cycle). Here, I_DC = 10 A and sinusoidal amplitude I_m = 10 A (sign does not matter for RMS).


Step-by-Step Solution:

Compute AC component RMS: I_AC,RMS = I_m/√2 = 10/√2 A.Square and add: I_RMS^2 = 10^2 + (10/√2)^2 = 100 + 50 = 150.Take square root: I_RMS = √150 A ≈ 12.247 A.


Verification / Alternative check:

Direct integral: I_RMS = sqrt{(1/T) ∫_0^T [10 − 10 sin(ωt)]^2 dt} yields the same result.


Why Other Options Are Wrong:

10 A is just the DC level; ignores the AC contribution.5 A and 7.07 A are typical individual components, not the combined RMS.10√2 A is the peak magnitude of the sum in some instants, not RMS.


Common Pitfalls:

Forgetting that RMS of DC + AC adds in quadrature; misreading sin(-t) as changing amplitude.


Final Answer:

√150 A (≈ 12.25 A)
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