What is the impedance of a circuit with a resistor of 4Ω and a capacitor of 2μF at a frequency of 50Hz? (2020)

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Q: What is the impedance of a circuit with a resistor of 4Ω and a capacitor of 2μF at a frequency of 50Hz? (2020)

Solution: Impedance Z = √(R² + (1/(ωC))²) where ω = 2πf. Here, R = 4Ω, C = 2μF, f = 50Hz. Calculate Z.

Steps: 9

Step 1: Identify the values given in the problem. We have a resistor (R) of 4Ω and a capacitor (C) of 2μF. The frequency (f) is 50Hz.
Step 2: Convert the capacitance from microfarads to farads. 2μF = 2 x 10^-6 F.
Step 3: Calculate the angular frequency (ω) using the formula ω = 2πf. Substitute f = 50Hz: ω = 2π(50).
Step 4: Calculate ω. This gives ω ≈ 314.16 rad/s.
Step 5: Calculate the capacitive reactance (Xc) using the formula Xc = 1/(ωC). Substitute ω and C: Xc = 1/(314.16 * 2 x 10^-6).
Step 6: Calculate Xc. This gives Xc ≈ 1591.55Ω.
Step 7: Use the impedance formula Z = √(R² + Xc²). Substitute R = 4Ω and Xc ≈ 1591.55Ω.
Step 8: Calculate Z. This gives Z ≈ √(4² + 1591.55²).
Step 9: Perform the calculation to find Z.