In an AC circuit operating at f = 5 MHz with V_rms = 1.0 V, a 27 pF capacitor is connected in parallel with a 1 kΩ resistor. What is the magnitude of the total current drawn from the 1 V_ac source?

Difficulty: Medium

Correct Answer: 1.31 mA

Explanation:


Introduction:
In a parallel RC circuit driven by a sinusoidal source, the resistor current and the capacitor current are not in phase. The resistor branch current is in phase with voltage, while the capacitor branch current leads the voltage by 90 degrees. The total source current is therefore the vector (phasor) sum of these orthogonal branch currents, not the arithmetic sum.


Given Data / Assumptions:

  • Supply voltage (rms): V = 1.0 V
  • Frequency: f = 5 MHz
  • Resistor: R = 1 kΩ
  • Capacitor: C = 27 pF
  • Ideal components, steady-state sinusoidal excitation


Concept / Approach:
For the resistor branch: I_R = V / R. For the capacitor branch: I_C = V / X_C, where X_C = 1 / (2 * π * f * C). Because the two currents are 90 degrees apart, the resultant magnitude is I_total = sqrt(I_R^2 + I_C^2).


Step-by-Step Solution:
Compute capacitive reactance: X_C = 1 / (2 * π * f * C)2 * π * f * C = 2 * π * 5e6 * 27e-12 ≈ 8.482e-4X_C ≈ 1 / 8.482e-4 ≈ 1179 ΩResistor current (rms): I_R = V / R = 1 / 1000 = 0.001 A = 1.0 mACapacitor current (rms): I_C = V / X_C ≈ 1 / 1179 ≈ 0.000848 A = 0.848 mATotal current magnitude: I_total = sqrt(I_R^2 + I_C^2) ≈ sqrt((1.0 mA)^2 + (0.848 mA)^2)I_total ≈ sqrt(1 + 0.719) mA ≈ sqrt(1.719) mA ≈ 1.31 mA


Verification / Alternative check:
Because branches are in parallel, voltages are equal. Orthogonal currents combine by Pythagoras. The computed X_C around 1.18 kΩ at 5 MHz for 27 pF is reasonable and results in I_C slightly less than I_R, so a total just above 1 mA is expected, agreeing with 1.31 mA.


Why Other Options Are Wrong:

  • 459 µA: Underestimates by ignoring the resistor current magnitude.
  • 647 µA: Roughly half the correct value; not consistent with vector addition.
  • 1.85 mA: Overestimates by incorrectly adding branch currents arithmetically.
  • 0.85 mA: This is close to I_C alone, ignoring I_R.


Common Pitfalls:

  • Adding branch currents directly instead of vectorially.
  • Using peak instead of rms values inconsistently.
  • Computing X_C with rad/s but plugging f in Hz incorrectly.


Final Answer:
1.31 mA

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