In a series L - C circuit, if `L = 10^(-3) H and C = 3xx10^(-7)` F is connected to a 100 V-50 Hz a.c.
Source, the impedance of the circuit is

Options

A. `(10^(5))/(3pi)-10pi`
B. `0.1pi -.3xx10^(-5)pi`
C. `(10^(5))/(3pi)-10pi`
D. None of these

Answer

To find the impedance of a series L-C circuit connected to an AC source, we can follow these steps:

### Step 1: Identify Given Values
- Inductance, \( L = 10^{-3} \, H \)
- Capacitance, \( C = 3 \times 10^{-7} \, F \)
- Frequency, \( f = 50 \, Hz \)

### Step 2: Calculate Angular Frequency
The angular frequency \( \omega \) is given by the formula:
\[ \omega = 2 \pi f \]
Substituting the given frequency:
\[ \omega = 2 \pi \times 50 = 100 \pi \, rad/s \]

### Step 3: Calculate Inductive Reactance \( X_L \)
Inductive reactance \( X_L \) is calculated using the formula:
\[ X_L = \omega L \]
Substituting the values:
\[ X_L = 100 \pi \times 10^{-3} = 0.1 \pi \, \Omega \]

### Step 4: Calculate Capacitive Reactance \( X_C \)
Capacitive reactance \( X_C \) is calculated using the formula:
\[ X_C = \frac{1}{\omega C} \]
Substituting the values:
\[ X_C = \frac{1}{100 \pi \times 3 \times 10^{-7}} = \frac{1}{3 \times 10^{-5} \pi} \, \Omega \]

### Step 5: Calculate Impedance \( Z \)
In a series L-C circuit, the impedance \( Z \) is given by:
\[ Z = \sqrt{(X_C - X_L)^2} \]
Since there is no resistance \( R = 0 \):
\[ Z = |X_C - X_L| \]
Substituting the values of \( X_C \) and \( X_L \):
\[ Z = \left| \frac{1}{3 \times 10^{-5} \pi} - 0.1 \pi \right| \]

### Step 6: Simplify and Calculate
To simplify, we need to find a common denominator:
\[ Z = \left| \frac{1 - 0.1 \pi \times 3 \times 10^{-5} \pi}{3 \times 10^{-5} \pi} \right| \]
Calculating \( 0.1 \pi \times 3 \times 10^{-5} \):
\[ 0.1 \times 3 \times 10^{-5} \pi = 3 \times 10^{-6} \pi \]
Thus:
\[ Z = \left| \frac{1 - 3 \times 10^{-6} \pi}{3 \times 10^{-5} \pi} \right| \]

### Final Calculation
Now, substituting the approximate value of \( \pi \approx 3.14 \):
\[ Z \approx \left| \frac{1 - 3 \times 10^{-6} \times 3.14}{3 \times 10^{-5} \times 3.14} \right| \]
Calculating the numerator:
\[ 1 - 9.42 \times 10^{-6} \approx 1 \]
Calculating the denominator:
\[ 3 \times 10^{-5} \times 3.14 \approx 9.42 \times 10^{-5} \]
Thus:
\[ Z \approx \frac{1}{9.42 \times 10^{-5}} \approx 10610 \, \Omega \]

### Conclusion
The impedance of the circuit is approximately \( Z \approx 10610 \, \Omega \).

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