For an alternating current of frequency `(500 )/(pi)` Hz in L-C-R series circuit with L = 1H, C = `1muF, R = 100 Omega`, impedance is-

Options

A. `100 Omega`
B. `100 sqrt(pi )Omega`
C. `100 sqrt(2pi) Omega`
D. `100 pi Omega`

Answer

To find the impedance (Z) in an L-C-R series circuit with given values, we can follow these steps:

### Step 1: Identify the given values
- Inductance (L) = 1 H
- Capacitance (C) = 1 µF = \(1 \times 10^{-6}\) F
- Resistance (R) = 100 Ω
- Frequency (f) = \(\frac{500}{\pi}\) Hz

### Step 2: Calculate the angular frequency (ω)
The angular frequency (ω) is given by the formula:
\[ \omega = 2\pi f \]
Substituting the value of f:
\[ \omega = 2\pi \left(\frac{500}{\pi}\right) = 1000 \text{ rad/s} \]

### Step 3: Calculate the inductive reactance (X_L)
Inductive reactance (X_L) is given by the formula:
\[ X_L = \omega L \]
Substituting the values:
\[ X_L = 1000 \times 1 = 1000 \, \Omega \]

### Step 4: Calculate the capacitive reactance (X_C)
Capacitive reactance (X_C) is given by the formula:
\[ X_C = \frac{1}{\omega C} \]
Substituting the values:
\[ X_C = \frac{1}{1000 \times 1 \times 10^{-6}} = \frac{1}{0.001} = 1000 \, \Omega \]

### Step 5: Calculate the impedance (Z)
The impedance in an L-C-R circuit is calculated using the formula:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
Since \(X_L\) and \(X_C\) are equal:
\[ Z = \sqrt{R^2 + (1000 - 1000)^2} = \sqrt{R^2} = R \]
Substituting the value of R:
\[ Z = \sqrt{100^2} = 100 \, \Omega \]

### Final Answer
The impedance (Z) of the circuit is \(100 \, \Omega\).

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