Openai/691e301d-9ef8-800a-b179-3a4e29e49fe1
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gpt-5-1[edit | edit source]
User: What is the impedance of an 0.1uF capacitor at 50Hz?[edit | edit source]
What is the impedance of an 0.1uF capacitor at 50Hz?
Assistant: To find the impedance of a capacitor, use:[edit | edit source]
To find the impedance of a capacitor, use:
ZC=1j 2πfCZ_C = \frac{1}{j \, 2\pi f C}ZC=j2πfC1 We usually give the magnitude:
∣ZC∣=12πfC|Z_C| = \frac{1}{2\pi f C}∣ZC∣=2πfC1 Given:
- C=0.1 μF=1×10−7 FC = 0.1\,\mu\text{F} = 1\times10^{-7}\,\text{F}C=0.1μF=1×10−7F
- f=50 Hzf = 50\,\text{Hz}f=50Hz
Compute:
∣ZC∣=12π⋅50⋅1×10−7|Z_C| = \frac{1}{2\pi \cdot 50 \cdot 1\times10^{-7}}∣ZC∣=2π⋅50⋅1×10−71 ∣ZC∣=13.14159×10−5|Z_C| = \frac{1}{3.14159 \times 10^{-5}}∣ZC∣=3.14159×10−51 ∣ZC∣≈31831 Ω|Z_C| \approx 31831\,\Omega∣ZC∣≈31831Ω So:
Impedance (magnitude): ~31.8 kΩ[edit | edit source]
Complex form:[edit | edit source]
ZC≈−j 31.8 kΩZ_C \approx -j\,31.8\,\text{k}\OmegaZC≈−j31.8kΩ (It's purely reactive and negative imaginary.)
References[edit | edit source]
- ↑ "Capacitor impedance calculation". ChatGPT. Retrieved 2025-11-20.