Question 3-16B5

To determine the impedance of parallel components, it's often easiest to work with admittance. 1. **Calculate Capacitive Reactance ($X_C$):** $X_C = 1 / (2\pi fC)$ $X_C = 1 / (2\pi \cdot 50,000 \text{ Hz} \cdot 0.01 \cdot 10^{-6} \text{ F})$ $X_C \approx 318.3 \Omega$ 2. **Calculate Admittance ($Y$) for each component:** * Resistive Admittance ($Y_R$) = $1/R = 1/300 \Omega \approx 0.00333 \text{ S}$ * Capacitive Admittance ($Y_C$) = $j/X_C = j/318.3 \Omega \approx j0.00314 \text{ S}$ 3. **Sum Total Admittance ($Y_{total}$):** For parallel components, admittances add directly. $Y_{total} = Y_R + Y_C = 0.00333 + j0.00314 \text{ S}$ 4. **Calculate Total Impedance ($Z$):** Impedance is the reciprocal of admittance. $Z = 1/Y_{total} = 1 / (0.00333 + j0.00314)$ To rationalize this complex number, multiply the numerator and denominator by the conjugate of the denominator: $Z = (1 \cdot (0.00333 - j0.00314)) / ((0.00333 + j0.00314) \cdot (0.00333 - j0.00314))$ $Z = (0.00333 - j0.00314) / (0.00333^2 + 0.00314^2)$ $Z \approx (0.00333 - j0.00314) / 0.00002095 \approx 159 - j150 \Omega$ The negative imaginary part ($-j$) correctly indicates a capacitive impedance. Options A, B, and D either have incorrect magnitudes or an incorrect sign for the imaginary component.