Question:

For the circuit shown, the locus of the impedance $Z(j\omega)$ is plotted as $\omega$ increases from zero to infinity. The values of $R_1$ and $R_2$ are:

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For $R \parallel C$ circuits, the Nyquist (impedance) plot forms a semicircle whose endpoints reveal $R_1$ and $R_1 + R_2$ directly.
Updated On: Dec 15, 2025
  • $R_1 = 2\ \text{k}\Omega,\ \ R_2 = 3\ \text{k}\Omega$
  • $R_1 = 5\ \text{k}\Omega,\ \ R_2 = 2\ \text{k}\Omega$
  • $R_1 = 5\ \text{k}\Omega,\ \ R_2 = 2.5\ \text{k}\Omega$
  • $R_1 = 2\ \text{k}\Omega,\ \ R_2 = 5\ \text{k}\Omega$
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The Correct Option is A

Solution and Explanation

The impedance is: \[ Z(j\omega) = R_1 + \left( R_2 \parallel \frac{1}{j\omega C} \right). \] At $\omega = 0$ (capacitor open-circuit), the parallel branch is: \[ Z(0) = R_1 + R_2. \] From the plot, $\Re(Z(0)) \approx 5\ \text{k}\Omega$. At $\omega = \infty$ (capacitor short-circuit), the parallel branch becomes: \[ Z(\infty) = R_1. \] From the plot, $\Re(Z(\infty)) \approx 2\ \text{k}\Omega$. Thus: \[ R_1 = 2\ \text{k}\Omega,\qquad R_2 = (5 - 2)\ \text{k}\Omega = 3\ \text{k}\Omega. \] These values also correctly match the semicircular locus shape for an $R_2 \parallel C$ branch added to $R_1$. Final Answer: $R_1 = 2\ \text{k}\Omega,\ R_2 = 3\ \text{k}\Omega$
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