Ohm’s law in AC magnitude: A resistor–capacitor series circuit has an impedance magnitude of 4.33 kΩ and is driven by a 20 V (rms) AC source. What is the rms current in the circuit?

Introduction / Context:
For any series AC circuit, the same current flows through all elements. With the magnitude of impedance |Z| known, Ohm’s law in phasor magnitude form I = V/|Z| immediately yields the rms current.

Given Data / Assumptions:

• |Z| = 4.33 kΩ = 4330 Ω.
• V_rms = 20 V.
• Sinusoidal steady state; use magnitudes.

Concept / Approach:

Apply I_rms = V_rms / |Z|. Units must be consistent (volts and ohms yield amperes). Convert to milliamperes for a practical figure.

Step-by-Step Solution:

Verification / Alternative check:

A 20 V source across several kilo-ohms should produce only a few milliamperes, which matches the calculated value.

Why Other Options Are Wrong:

9.2 mA doubles the correct value; 92 mA and 460 mA are far too large for 4.33 kΩ at 20 V.

Common Pitfalls:

Forgetting to convert kΩ to Ω; misplacing decimals when converting to mA.

Final Answer:

4.6 mA