For the parallel combination of a 0.01 µF capacitor and a 300-ohm resistor at 50 kHz, which value is closest to the real part of the impedance?

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Multiple Choice

For the parallel combination of a 0.01 µF capacitor and a 300-ohm resistor at 50 kHz, which value is closest to the real part of the impedance?

In a parallel RC, the total impedance is complex because the capacitor adds a reactive path. The real part of the parallel impedance can be found from the relationship for a resistor R in parallel with a capacitor C: Re{Z_eq} = R / [1 + (ω R C)²], where ω = 2πf. Here, C = 0.01 µF = 1×10⁻⁸ F, f = 50 kHz, so ω = 2π×50,000 ≈ 3.1416×10⁵ rad/s. Compute ωRC: ωRC = (3.1416×10⁵) × 300 × (1×10⁻⁸) ≈ 0.9425. Then Re{Z_eq} = 300 / [1 + (0.9425)²] ≈ 300 / [1 + 0.888] ≈ 300 / 1.888 ≈ 159 Ω. So the real part is about 159 Ω. The parallel combination also has an imaginary part (capacitive) of roughly −150 Ω, giving Z_eq ≈ 159 − j150 Ω. This explains why the real part isn’t simply the resistor value; the capacitor’s presence reduces the effective resistive component at this frequency.

In a parallel RC, the total impedance is complex because the capacitor adds a reactive path. The real part of the parallel impedance can be found from the relationship for a resistor R in parallel with a capacitor C: Re{Z_eq} = R / [1 + (ω R C)²], where ω = 2πf.

Here, C = 0.01 µF = 1×10⁻⁸ F, f = 50 kHz, so ω = 2π×50,000 ≈ 3.1416×10⁵ rad/s. Compute ωRC: ωRC = (3.1416×10⁵) × 300 × (1×10⁻⁸) ≈ 0.9425. Then

Re{Z_eq} = 300 / [1 + (0.9425)²] ≈ 300 / [1 + 0.888] ≈ 300 / 1.888 ≈ 159 Ω.

So the real part is about 159 Ω. The parallel combination also has an imaginary part (capacitive) of roughly −150 Ω, giving Z_eq ≈ 159 − j150 Ω. This explains why the real part isn’t simply the resistor value; the capacitor’s presence reduces the effective resistive component at this frequency.

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