Question 3-17B1

To determine the impedance of a parallel RC network, first calculate the capacitive reactance ($X_C$) at the given frequency: $X_C = \frac{1}{2 \pi f C}$ Where $f = 500 \text{ KHz} = 500 \times 10^3 \text{ Hz}$ and $C = 100 \text{ pF} = 100 \times 10^{-12} \text{ F}$. $X_C = \frac{1}{2 \pi (500 \times 10^3 \text{ Hz}) (100 \times 10^{-12} \text{ F})} \approx 3183.1 \text{ ohms}$. For a parallel circuit, it's often easier to work with admittances ($Y$), where $Y = \frac{1}{Z}$. The admittance of the resistor is $Y_R = \frac{1}{R} = \frac{1}{4000 \text{ ohms}} = 0.00025 \text{ S}$. The admittance of the capacitor is $Y_C = \frac{1}{-j X_C} = j \frac{1}{X_C} = j \frac{1}{3183.1 \text{ ohms}} \approx j 0.000314159 \text{ S}$. The total admittance $Y_{total}$ is the sum of the individual admittances: $Y_{total} = Y_R + Y_C = 0.00025 + j 0.000314159 \text{ S}$. To convert $Y_{total}$ to polar form: Magnitude $|Y_{total}| = \sqrt{(0.00025)^2 + (0.000314159)^2} \approx 0.0004015 \text{ S}$. Phase angle $\theta_Y = \arctan\left(\frac{0.000314159}{0.00025}\right) \approx 51.5^\circ$. So, $Y_{total} \approx 0.0004015 \angle 51.5^\circ \text{ S}$. Finally, the total impedance $Z_{total}$ is the reciprocal of the total admittance: $Z_{total} = \frac{1}{Y_{total}} = \frac{1}{0.0004015 \angle 51.5^\circ} \text{ ohms}$. Magnitude $|Z_{total}| = \frac{1}{0.0004015} \approx 2490 \text{ ohms}$. Phase angle $\theta_Z = -51.5^\circ$ (the angle of impedance is the negative of the angle of admittance). Thus, the impedance is 2490 ohms, /-51.5 degrees. Option A has the correct magnitude but the wrong phase angle (positive instead of negative for a capacitive circuit). Options B and C have incorrect magnitudes and angles. The final answer is $\boxed{D}$