In a parallel network consisting of a 100 Ω resistor and a capacitor with reactance 100 Ω at the operating frequency, what is the phase angle between the total current and the applied voltage?

Introduction / Context:Parallel RC circuits require phasor reasoning because branch currents are out of phase. This question checks understanding of how resistive and capacitive branch currents combine vectorially to produce a total current that leads the applied voltage by a specific angle.

Given Data / Assumptions:

• Resistor R = 100 Ω in parallel with a capacitor of reactance Xc = 100 Ω.
• Steady state sinusoidal excitation; ideal components.
• We seek the magnitude of the phase shift between total current and voltage.

Concept / Approach:In parallel RC, I_R = V / R is in phase with V and I_C = V / Xc leads V by 90 degree. The total current I_T is the phasor sum of I_R and I_C with a right angle between them. The phase angle theta by which I_T leads V is theta = arctan( I_C / I_R ) = arctan( (V/Xc) / (V/R) ) = arctan( R / Xc ). For R = Xc, theta = arctan(1) = 45 degree.

Step-by-Step Solution:Step 1: Compute I_R = V / R.Step 2: Compute I_C = V / Xc with a +90 degree phase relative to V.Step 3: Determine theta = arctan( I_C / I_R ) = arctan( R / Xc ).Step 4: Substitute R = 100 Ω and Xc = 100 Ω to get theta = 45 degree lead.

Verification / Alternative check:Phasor diagram forms a right triangle with equal legs when R = Xc; the angle with the in-phase axis is 45 degree, confirming the result.

Why Other Options Are Wrong:

• 180 degree: Would indicate inversion, not applicable here.
• 30 degree or 75 degree: Do not match the equal-leg right triangle.
• None of the above: Incorrect because 45 degree is correct.

Common Pitfalls:Adding branch currents arithmetically ignores phase and gives wrong angles. Always use vector addition or compute theta with arctan of the ratio of quadrature and in-phase components.

Final Answer:45 degree.