A resistor of 100 ohm is connected in series with an inductor of 10H and a capacitor of `0.1 mu F` All these elements are connected to a 220 volt, 50 Hz a.c. supply. Calculate the total impedance of the circuit.

Answer

To calculate the total impedance of the circuit consisting of a resistor, inductor, and capacitor connected in series, we can follow these steps:

### Step 1: Identify the given values
- Resistance \( R = 100 \, \Omega \)
- Inductance \( L = 10 \, H \)
- Capacitance \( C = 0.1 \, \mu F = 0.1 \times 10^{-6} \, F = 10^{-7} \, F \)
- Frequency \( f = 50 \, Hz \)

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

### Step 3: Calculate the inductive reactance \( X_L \)
The inductive reactance \( X_L \) is calculated using the formula:
\[ X_L = \omega L \]
Substituting the values:
\[ X_L = 100 \pi \times 10 = 1000 \pi \, \Omega \]

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

### Step 5: Calculate the total impedance \( Z \)
The total impedance \( Z \) for a series circuit is given by:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
Substituting the values:
\[ Z = \sqrt{100^2 + (1000 \pi - \frac{10^5}{\pi})^2} \]
Calculating \( R^2 \):
\[ R^2 = 100^2 = 10000 \]
Calculating \( X_L - X_C \):
\[ X_L - X_C = 1000 \pi - \frac{10^5}{\pi} \]
Now squaring this term:
\[ (X_L - X_C)^2 = (1000 \pi - \frac{10^5}{\pi})^2 \]
Now substituting back into the impedance formula:
\[ Z = \sqrt{10000 + (1000 \pi - \frac{10^5}{\pi})^2} \]

### Step 6: Calculate the final value of \( Z \)
To simplify the calculations:
1. Calculate \( 1000 \pi \approx 3141.59 \)
2. Calculate \( \frac{10^5}{\pi} \approx 31830.99 \)
3. Now \( X_L - X_C \approx 3141.59 - 31830.99 \approx -28689.4 \)
4. Square this value and add to \( R^2 \):
\[ Z \approx \sqrt{10000 + (-28689.4)^2} \]
Calculating this gives:
\[ Z \approx \sqrt{10000 + 822000000} \approx \sqrt{822010000} \approx 28683.5 \, \Omega \]

### Final Answer:
The total impedance \( Z \) of the circuit is approximately \( 28683.5 \, \Omega \).

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