A voltage $V_{PQ}=V_0\cos\omega t$ is applied between the points P and Q in the network shown. $C=\frac{1}{\omega R\sqrt{3}}$ and $L=\frac{R\sqrt{3}}{\omega}$. The total impedance between P and Q is
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In AC circuits, when an inductive branch and a capacitive branch have identical resistance and reactance magnitudes, the parallel combination becomes purely resistive. In this case, the complex parts cancel out, simplifying the calculation to $Z_p = \frac{R^2 + X^2}{2R}$.
Concept:
The circuit consists of a resistor $R$ at the input (point P) connected in series with a parallel combination of an RL branch and an RC branch.
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• $X_L = \omega L = \omega \left( \frac{R\sqrt{3}}{\omega} \right) = R\sqrt{3}$
• $X_C = \frac{1}{\omega C} = \frac{1}{\omega \left( \frac{1}{\omega R \sqrt{3}} \right)} = R\sqrt{3}$
Step 1: Calculate impedance of the parallel branches.
Upper branch: $Z_1 = R + jX_L = R + jR\sqrt{3}$.
Lower branch: $Z_2 = R - jX_C = R - jR\sqrt{3}$.
Parallel impedance $Z_p$:
\[ Z_p = \frac{Z_1 Z_2}{Z_1 + Z_2} = \frac{(R + jR\sqrt{3})(R - jR\sqrt{3})}{(R + jR\sqrt{3}) + (R - jR\sqrt{3})} \]
\[ Z_p = \frac{R^2 - (jR\sqrt{3})^2}{2R} = \frac{R^2 + 3R^2}{2R} = \frac{4R^2}{2R} = 2R \]
Step 2: Calculate total impedance $Z_{PQ}$.
The resistor $R$ connected to point P is in series with the parallel combination $Z_p$ calculated above:
\[ Z_{PQ} = R + Z_p \]
\[ Z_{PQ} = R + 2R = 3R \]
