Parallel RLC phasor analysis: Given branch currents IL = 15.3 A (inductive branch, current lags voltage by 90°), IC = 0.43 A (capacitive branch, current leads voltage by 90°), and IR = 3.5 A (resistive branch, in phase with voltage), determine the circuit’s overall current phase angle θ in degrees (negative for inductive).

Difficulty: Medium

Correct Answer: -76.7 degrees

Explanation:


Introduction:
In a parallel RLC circuit, the total current is the phasor (vector) sum of branch currents: resistive current IR is in phase with the supply voltage, inductive current IL lags the voltage by 90 degrees, and capacitive current IC leads the voltage by 90 degrees. The net phase angle of the total current with respect to the voltage reveals whether the circuit is overall inductive (negative angle, current lags) or capacitive (positive angle, current leads).


Given Data / Assumptions:

  • Inductive branch current IL = 15.3 A (lags V by 90 degrees)
  • Capacitive branch current IC = 0.43 A (leads V by 90 degrees)
  • Resistive branch current IR = 3.5 A (in phase with V)
  • Sinusoidal steady state; ideal components


Concept / Approach:

Let the in-phase axis be along the voltage (and IR). The reactive axis is orthogonal. Net reactive current is Ireact = IC - IL (capacitive positive, inductive negative). The phase angle of total current is θ = arctan(Ireact / IR). A negative θ means the total current lags the voltage (inductive dominance).


Step-by-Step Solution:

Compute net reactive current: Ireact = IC - IL = 0.43 - 15.3 = -14.87 AUse θ = arctan(Ireact / IR)θ = arctan((-14.87) / 3.5) ≈ arctan(-4.2486)θ ≈ -76.7 degrees (inductive overall)


Verification / Alternative check:

The magnitude of the total current is |I| = sqrt(IR^2 + Ireact^2) ≈ sqrt(3.5^2 + 14.87^2) ≈ 15.3 A. Since |Ireact| > IR, the angle must be near ±90 degrees; negative sign confirms an inductive net response.


Why Other Options Are Wrong:

  • 76.7 degrees: Wrong sign; would imply strong capacitive lead.
  • -4.25 degrees: Magnitude far too small given |Ireact| ≫ IR.
  • 88.8 degrees: Too large; |Ireact|/IR ≈ 4.25, not near infinity.
  • -15.0 degrees: Underestimates inductive dominance.


Common Pitfalls:

  • Adding magnitudes arithmetically instead of using phasors.
  • Forgetting that Ireact = IC - IL with sign convention (capacitive positive).


Final Answer:

-76.7 degrees

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