If resistance R =100 Omega, inductance L = 12mH and capacitance C=15muF are connected in series to an AC sources, then at resonance the impedance of circuit is

Options

A. zero
B. 100Omega
C. 1000Omega
D. 10kOmega

Answer

To solve for the impedance of a series RLC circuit at resonance, we will follow these steps:

### Step 1: Understand the Components
We are given:
- Resistance \( R = 100 \, \Omega \)
- Inductance \( L = 12 \, \text{mH} = 12 \times 10^{-3} \, \text{H} \)
- Capacitance \( C = 15 \, \mu\text{F} = 15 \times 10^{-6} \, \text{F} \)

### Step 2: Resonance Condition
At resonance in an RLC circuit, the inductive reactance \( X_L \) is equal to the capacitive reactance \( X_C \):
\[ X_L = X_C \]
Where:
- Inductive reactance \( X_L = \omega L \)
- Capacitive reactance \( X_C = \frac{1}{\omega C} \)

### Step 3: Set Up the Resonance Equation
At resonance:
\[ \omega L = \frac{1}{\omega C} \]
This implies:
\[ \omega^2 = \frac{1}{LC} \]

### Step 4: Calculate \( \omega \)
Substituting the values of \( L \) and \( C \):
\[ \omega^2 = \frac{1}{(12 \times 10^{-3})(15 \times 10^{-6})} \]
Calculating \( LC \):
\[ LC = 12 \times 10^{-3} \times 15 \times 10^{-6} = 180 \times 10^{-9} \, \text{H}\cdot\text{F} \]
Thus:
\[ \omega^2 = \frac{1}{180 \times 10^{-9}} \approx 5.56 \times 10^6 \]
Taking the square root:
\[ \omega \approx 2355 \, \text{rad/s} \]

### Step 5: Calculate \( X_L \) and \( X_C \)
Now we can find \( X_L \) and \( X_C \):
\[ X_L = \omega L = 2355 \times 12 \times 10^{-3} \approx 28.26 \, \Omega \]
\[ X_C = \frac{1}{\omega C} = \frac{1}{2355 \times 15 \times 10^{-6}} \approx 28.26 \, \Omega \]

### Step 6: Calculate Total Impedance \( Z \)
At resonance, since \( X_L = X_C \), the total impedance \( Z \) is simply the resistance \( R \):
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{100^2 + (28.26 - 28.26)^2} = \sqrt{100^2} = 100 \, \Omega \]

### Final Answer
Thus, the impedance of the circuit at resonance is:
\[ Z = 100 \, \Omega \]