Question

In a series RLC circuit, if the frequency of the applied voltage is increased, the impedance of the circuit:

• A. Increases
• B. Decreases
• C. Remains constant
• D. First increases then decreases

Hint:

The impedance in a series RLC circuit decreases as the frequency increases towards resonance. At resonance, the impedance is at its minimum and is equal to the resistance \( R \). After resonance, the impedance starts to increase again.

Answer 1

The Correct Option is B

In a series RLC circuit, the total impedance \( Z \) is given by: \[ Z = \sqrt{R^2 + (X_L - X_C)^2} \] where:
\( R \) is the resistance,
\( X_L = 2\pi f L \) is the inductive reactance,
\( X_C = \frac{1}{2\pi f C} \) is the capacitive reactance,
\( f \) is the frequency of the applied voltage,
\( L \) is the inductance, and
\( C \) is the capacitance.
Step 1: At very low frequencies, when \( f \) is small, the capacitive reactance \( X_C \) dominates because it is inversely proportional to the frequency (\( X_C = \frac{1}{2\pi f C} \)). So, the impedance is high. 
Step 2: As the frequency increases, the inductive reactance \( X_L \) increases (since \( X_L = 2\pi f L \)) and the capacitive reactance \( X_C \) decreases. 
Step 3: At the resonant frequency (\( f_0 \)), the inductive reactance and capacitive reactance are equal, \( X_L = X_C \), so the impedance reaches its minimum value, which is just the resistance \( Z = R \). 
Step 4: If the frequency is increased further past the resonance point, the inductive reactance \( X_L \) becomes dominant, and the impedance increases again. But in the range before resonance, the impedance decreases as the frequency increases. Thus, before resonance (as the frequency increases from a low value to the resonant frequency), the impedance of the series RLC circuit decreases. Therefore, the correct answer is (b) Decreases.