Capacitive reactance calculation — with a 5 V, 60 Hz source applied to a 4.7 µF capacitor, what is the circuit current (assume ideal components and RMS values)?

Introduction / Context:
For a pure capacitor driven by a sinusoidal voltage source, the magnitude of the current is determined by the capacitive reactance Xc. This problem tests the ability to compute Xc and then use Ohm’s law in the phasor (RMS) sense to find current.

Given Data / Assumptions:

• V_rms = 5 V (sinusoidal, 60 Hz).
• Capacitance C = 4.7 µF.
• Ideal capacitor, no series resistance; RMS quantities used.

Concept / Approach:
Capacitive reactance magnitude is Xc = 1 / (2 * pi * f * C). The current magnitude then is I_rms = V_rms / Xc for a purely reactive element. After computing Xc, divide 5 V by Xc to get the current in amperes and convert to milliamperes.

Step-by-Step Solution:

Verification / Alternative check:
A quick sanity check: increasing C lowers Xc and raises current; for a few microfarads at 60 Hz, a few milliamps is typical at low voltages, which agrees with 8.86 mA at 5 V.

Why Other Options Are Wrong:

• 0.886 mA / 88.6 µA / 8.86 µA: Each is off by one or two decimal places relative to the computed value.

Common Pitfalls:
Mixing peak and RMS values; misplacing decimal points when converting microfarads to farads; forgetting that capacitive current leads voltage by 90 degrees (phase does not change magnitude here).

Final Answer:
8.86 mA.