A resistance (R) = 12 Omega, inductance (L) = 2 henry and capacitive reactance C = 5 mF are connected in series to an ac generator

Options

A. at resonance, the circuit impedance is zero.
B. at resonance, the circuit impedance is 12Omega
C. the resonance frequency of the circuit is 1//2pi.
D. at resonance, the inductive reactance is less than the capacitive reactance

Answer

To solve the problem step by step, we will analyze the given components in the AC circuit and determine the correct options based on the conditions at resonance.

### Step 1: Identify the given values
- Resistance (R) = 12 Ω
- Inductance (L) = 2 H
- Capacitance (C) = 5 mF = 5 × 10^-3 F

### Step 2: Calculate the inductive reactance (XL)
The inductive reactance (XL) is given by the formula:
\[ X_L = 2 \pi f L \]
However, we need to find the resonance frequency first, so we will calculate it later.

### Step 3: Calculate the capacitive reactance (XC)
The capacitive reactance (XC) is given by the formula:
\[ X_C = \frac{1}{2 \pi f C} \]

### Step 4: Condition for resonance
At resonance, the inductive reactance (XL) is equal to the capacitive reactance (XC):
\[ X_L = X_C \]

### Step 5: Calculate the impedance (Z) at resonance
The total impedance (Z) in a series RLC circuit is given by:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
At resonance, since \(X_L = X_C\), the impedance simplifies to:
\[ Z = R \]
Thus, in this case:
\[ Z = 12 \, \Omega \]

### Step 6: Analyze the options
1. **At resonance, the circuit impedance is zero.** - Incorrect, as we found \(Z = R\).
2. **At resonance, the circuit impedance is 12 Ω.** - Correct, as we found \(Z = 12 \, \Omega\).
3. **The resonance frequency of the circuit is \( \frac{1}{12 \pi} \).** - We will calculate the resonance frequency next.
4. **At resonance, the inductive reactance is less than the capacitive reactance.** - Incorrect, as at resonance \(X_L = X_C\).

### Step 7: Calculate the resonance frequency (f)
The resonance frequency (f) is given by:
\[ f = \frac{1}{2 \pi \sqrt{LC}} \]
Substituting the values:
\[ f = \frac{1}{2 \pi \sqrt{2 \times 5 \times 10^{-3}}} \]
Calculating:
\[ f = \frac{1}{2 \pi \sqrt{0.01}} = \frac{1}{2 \pi \times 0.1} = \frac{1}{0.2 \pi} = \frac{5}{\pi} \]
This is not equal to \( \frac{1}{12 \pi} \), so this option is incorrect.

### Final Summary of Options
- The correct option is that at resonance, the circuit impedance is 12 Ω.