If resistance R `= 10 Omega`, inductance L = 2 mH and capacitance `C = 5 muF` are connected in series to an AC source of frequency 50 Hz, then at resonance the impedance of circuit is

Options

A. Zero
B. `10 Omega`
C. `1000 Omega`
D. `10 K Omega`

Answer

To solve the problem, we need to determine the impedance of a series RLC circuit at resonance. Here’s a step-by-step solution:

### Step 1: Understand the Resonance Condition
At resonance in a series RLC circuit, the inductive reactance (XL) is equal to the capacitive reactance (XC). This means:
\[ XL = XC \]

### Step 2: Calculate Inductive Reactance (XL)
The inductive reactance is given by the formula:
\[ XL = \omega L \]
where \( \omega = 2\pi f \) and \( f \) is the frequency in hertz.

Given:
- \( L = 2 \, \text{mH} = 2 \times 10^{-3} \, \text{H} \)
- \( f = 50 \, \text{Hz} \)

First, calculate \( \omega \):
\[ \omega = 2\pi \times 50 = 100\pi \, \text{rad/s} \]

Now calculate \( XL \):
\[ XL = (100\pi)(2 \times 10^{-3}) = 0.2\pi \, \text{ohm} \]

### Step 3: Calculate Capacitive Reactance (XC)
The capacitive reactance is given by the formula:
\[ XC = \frac{1}{\omega C} \]

Given:
- \( C = 5 \, \mu F = 5 \times 10^{-6} \, F \)

Now calculate \( XC \):
\[ XC = \frac{1}{100\pi \times 5 \times 10^{-6}} = \frac{1}{0.0005\pi} \approx \frac{2000}{\pi} \, \text{ohm} \]

### Step 4: Set XL Equal to XC
At resonance:
\[ XL = XC \]
This gives us the condition for resonance.

### Step 5: Calculate Impedance (Z) at Resonance
The total impedance (Z) in a series RLC circuit at resonance is simply equal to the resistance (R):
\[ Z = R \]

Given:
- \( R = 10 \, \Omega \)

Thus, the impedance at resonance is:
\[ Z = 10 \, \Omega \]

### Final Answer
The impedance of the circuit at resonance is \( 10 \, \Omega \).

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