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In a series LCR circuit driven by an AC source, the current amplitude I is plotted against the angular frequency omega. The resonant angular frequency is omega_r. Pick the correct statement about this curve and circuit.

1. (a) For omega below omega_r, the circuit behaves mainly as a capacitive circuit.
2. (b) For omega below omega_r, the circuit behaves mainly as an inductive circuit.
3. (c) At omega = omega_r, the net impedance of the circuit equals just its resistance R.
4. (d) At omega = omega_r, the net impedance of the circuit is zero.

Correct answer: (c) At omega = omega_r, the net impedance of the circuit equals just its resistance R.

Solution

In a series LCR circuit, Z = sqrt(R² + (X_L - X_C)²) where X_L = omega*L and X_C = 1/(omega*C). At resonance omega_r = 1/sqrt(L*C), X_L = X_C, so the reactive parts cancel and Z = R (minimum impedance, maximum current). For omega < omega_r, X_C > X_L, so the circuit is net capacitive (not inductive). The impedance at resonance is never zero because R remains.

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