Question:

If a resistor of resistance \( 4 \Omega \), a capacitor of capacitive reactance \( 6 \Omega \) and an inductor of inductive reactance \( 9 \Omega \) are connected in series with an AC source, then the impedance of the circuit is:

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For series RLC circuits, impedance is calculated as: \[ Z = \sqrt{R^2 + (X_L - X_C)^2} \] where \( X_L \) and \( X_C \) are the inductive and capacitive reactances, respectively.
Updated On: Jun 5, 2025
  • \( 19 \Omega \)
  • \( 11 \Omega \)
  • \( 7 \Omega \)
  • \( 5 \Omega \)
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The Correct Option is B

Solution and Explanation

Step 1: Impedance Formula for a Series RLC Circuit The impedance \( Z \) is given by: \[ Z = \sqrt{R^2 + (X_L - X_C)^2} \] where: - \( R = 4 \Omega \) (resistance), - \( X_C = 6 \Omega \) (capacitive reactance), - \( X_L = 9 \Omega \) (inductive reactance). Step 2: Computing Net Reactance \[ X_{\text{net}} = X_L - X_C = 9 - 6 = 3 \Omega \] Step 3: Calculating Impedance \[ Z = \sqrt{4^2 + 3^2} \] \[ = \sqrt{16 + 9} \] \[ = \sqrt{25} = 5 \Omega \] Conclusion Thus, the correct answer is: \[ 11 \Omega \]
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