A resistance of 40 `Omega` is connected in series with inductor of self-inductance 5 H and a capacitor of capacitance 80 `mu`F. This combination is connected to an AC source of rms voltage 220 V. frequency of AC source can changed continuously.
What is the impedance of circuit in a state of resonance ?

Options

A. 40 `Omega`
B. 80 `Omega`
C. 400 `Omega`
D. 800 `Omega`

Explanation

To find the impedance of the circuit in a state of resonance, we can follow these steps:

### Step 1: Understand the Resonance Condition
In a series LCR circuit, resonance occurs when the inductive reactance (XL) is equal to the capacitive reactance (XC). At this point, the impedance (Z) of the circuit is minimized.

### Step 2: Write the Expression for Impedance
The impedance (Z) of a series LCR circuit is given by the formula:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
Where:
- \( R \) is the resistance,
- \( X_L \) is the inductive reactance,
- \( X_C \) is the capacitive reactance.

### Step 3: Set Up the Reactance Values
The inductive reactance \( X_L \) is given by:
\[ X_L = \omega L \]
And the capacitive reactance \( X_C \) is given by:
\[ X_C = \frac{1}{\omega C} \]
At resonance, we have:
\[ X_L = X_C \]
Thus,
\[ \omega L = \frac{1}{\omega C} \]

### Step 4: Solve for Impedance at Resonance
At resonance, since \( X_L = X_C \), the term \( (X_L - X_C) \) becomes zero. Therefore, the impedance simplifies to:
\[ Z = \sqrt{R^2 + 0^2} = R \]

### Step 5: Substitute the Given Resistance Value
Given that the resistance \( R = 40 \, \Omega \):
\[ Z = R = 40 \, \Omega \]

### Final Answer
Thus, the impedance of the circuit in a state of resonance is:
\[ \boxed{40 \, \Omega} \]

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