In a series R-C AC circuit, how does the capacitor affect impedance and phase between voltage and current?

• A. Impedance is Z = sqrt(R^2 + (1/(ωC))^2), and current leads voltage by φ = arctan(1/(ωRC)).
• B. Impedance is Z = R + 1/(ωC); current lags voltage by φ = arctan(ωRC).
• C. Impedance is Z = sqrt(R^2 + (ωC)^2); current is in phase with voltage.
• D. Impedance is Z = R + 1/(ωC); current leads by φ = arctan(1/(ωRC)).

Answer: (D)

In a series RC circuit, the capacitor introduces a reactive component that is 90 degrees out of phase with the resistor. This makes the total impedance a complex quantity: Z = R − j(1/ωC). The negative imaginary part means the current, which is proportional to the voltage divided by Z, will lead the voltage.

The magnitude of the impedance is the length of that phasor: |Z| = sqrt(R^2 + (1/(ωC))^2). The phase angle of the impedance relative to the voltage is θ = arctan(Im(Z)/Re(Z)) = arctan(−1/(ωCR)) = −arctan(1/(ωCR)). Since the impedance has a negative imaginary component, the voltage lags the current by that angle, so the current leads the voltage by φ = arctan(1/(ωRC)).

Context: increasing frequency reduces the capacitor’s reactance (Xc = 1/(ωC)), so the impedance becomes more like a pure resistor and the phase shift approaches zero. Decreasing frequency increases Xc, making the impedance more reactive and pushing the phase shift closer to ±90 degrees (in this case toward −90 degrees for the impedance, with the current leading by a larger amount).

So, the capacitor causes the current to lead the voltage by φ = arctan(1/(ωRC)), and the impedance magnitude is |Z| = sqrt(R^2 + (1/(ωC))^2).