A sinusoidal voltage of peak value 210 V and angular frequency 300/s is applied to a series LCR circuit. Given that `R-5Omega, L-25` mH and `C - 1000 muF`.
Calculate the total impedance of the circuit.

Options

A. `6.5 Omega`
B. `5 Omega`
C. `6 Omega`
D. `7 Omega`

Explanation

To calculate the total impedance of the given LCR circuit, we will follow these steps:

### Step 1: Calculate the Inductive Reactance (XL)
The formula for inductive reactance is given by:
\[ X_L = \omega L \]
Where:
- \(\omega = 300 \, \text{rad/s}\)
- \(L = 25 \, \text{mH} = 25 \times 10^{-3} \, \text{H}\)

Substituting the values:
\[ X_L = 300 \times (25 \times 10^{-3}) = 300 \times 0.025 = 7.5 \, \Omega \]

### Step 2: Calculate the Capacitive Reactance (XC)
The formula for capacitive reactance is given by:
\[ X_C = \frac{1}{\omega C} \]
Where:
- \(C = 1000 \, \mu F = 1000 \times 10^{-6} \, F\)

Substituting the values:
\[ X_C = \frac{1}{300 \times (1000 \times 10^{-6})} = \frac{1}{300 \times 0.001} = \frac{1}{0.3} \approx 3.33 \, \Omega \]

### Step 3: Calculate the Total Impedance (Z)
The total impedance in a series LCR circuit is given by:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
Where:
- \(R = 5 \, \Omega\)

Substituting the values:
\[ Z = \sqrt{5^2 + (7.5 - 3.33)^2} \]
Calculating \(X_L - X_C\):
\[ X_L - X_C = 7.5 - 3.33 = 4.17 \]
Now substituting this back into the impedance formula:
\[ Z = \sqrt{25 + (4.17)^2} \]
Calculating \((4.17)^2\):
\[ (4.17)^2 \approx 17.39 \]
Now substituting this value:
\[ Z = \sqrt{25 + 17.39} = \sqrt{42.39} \]
Calculating the square root:
\[ Z \approx 6.5 \, \Omega \]

### Final Answer:
The total impedance of the circuit is approximately \(6.5 \, \Omega\).

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