In a series R-L-C circuit at resonance, impedance is----

Analyze the question: We're asked what happens to the impedance of a series R-L-C circuit when it is at resonance.

Concept: The impedance of a series R-L-C circuit is:

Z = √[R² + (X_L − X_C)²]

where X_L = ωL (inductive reactance) and X_C = 1/(ωC) (capacitive reactance).

Solution:

1. Resonance occurs when the inductive reactance equals the capacitive reactance: X_L = X_C.
2. This gives X_L − X_C = 0.
3. The impedance becomes Z = √(R² + 0) = R.
4. Since the reactive part is zero, Z cannot get any smaller than R, so the impedance is at its minimum value.
5. As a result, the current is at its maximum, since I = V/Z = V/R.
6. The circuit behaves as a purely resistive circuit, so it is neither capacitive nor inductive. This rules out C and D.
7. Option A is wrong because maximum impedance happens in a parallel LC circuit at resonance, not a series one.

The resonant frequency is f₀ = 1/(2π√(LC)).

Answer: The correct option is B. minimum.

Exam tip: Remember "Series = Small impedance, Smax current". In a series resonance circuit, Z is minimum and I is maximum. In a parallel resonance circuit, it's the opposite: Z is maximum and I is minimum.