Question

In a series LCR Circuit R = 300 Ω, L = 0.9 H, c = 2.0 µF and w = 1000 rad/sec. Then the impedance of the circuit is 

• A. 500 Ω
• B. 1300 Ω
• C. 400 Ω
• D. 900 Ω

Answer 1

The Correct Option is A

The impedance of a series LCR circuit can be calculated using the formula:

\( Z = \sqrt{R^2 + (X_L - X_C)^2}\)

Where \( R \) is the resistance, \( X_L \) is the inductive reactance, and \( X_C \) is the capacitive reactance.

Given values:

• R = 300 Ω
• L = 0.9 H
• C = 2.0 µF
• ω = 1000 rad/sec

The inductive reactance \( X_L \) can be calculated as follows:

\( X_L = \omega L\)

Substituting the given values:

\( X_L = (1000 \, \text{rad/sec})(0.9 \, \text{H}) = 900 \, \Omega\)

The capacitive reactance \( X_C \) can be calculated as follows:

\( X_C = \frac{1}{\omega C}\)

Substituting the given values:

\( X_C = \frac{1}{(1000 \, \text{rad/sec})(2.0 \, \mu \text{F})} = \frac{1}{2 \times 10^{-6}} \, / \, 1000 = 500 \, \Omega\)

Now, we can substitute the values of \( R \), \( X_L \), and \( X_C \) into the impedance formula:

\( Z = \sqrt{300^2 + (900 - 500)^2} = \sqrt{90000 + 160000} = \sqrt{250000} = 500 \, \Omega\)

Therefore, the impedance of the circuit is 500 Ω. So the correct option is (A) 500 Ω.

Answer 2

The impedance \(Z\) of a series LCR circuit is given by:

\(Z = \sqrt{R^2 + (X_L - X_C)^2}\)

Where:

• \(R\) is the resistance
• \(X_L\) is the inductive reactance, \(X_L = \omega L\)
• \(X_C\) is the capacitive reactance, \(X_C = \frac{1}{\omega C}\)
• \(\omega\) is the angular frequency

Given:

• \(R = 300 \, \Omega\)
• \(L = 0.9 \, \text{H}\)
• \(C = 2.0 \, \mu\text{F} = 2.0 \times 10^{-6} \, \text{F}\)
• \(\omega = 1000 \, \text{rad/s}\)

Calculate the inductive reactance:

\(X_L = \omega L = (1000 \, \text{rad/s})(0.9 \, \text{H}) = 900 \, \Omega\)

Calculate the capacitive reactance:

\(X_C = \frac{1}{\omega C} = \frac{1}{(1000 \, \text{rad/s})(2.0 \times 10^{-6} \, \text{F})} = \frac{1}{0.002} = 500 \, \Omega\)

Calculate the impedance:

\(Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{(300)^2 + (900 - 500)^2}\)

\(Z = \sqrt{(300)^2 + (400)^2} = \sqrt{90000 + 160000} = \sqrt{250000} = 500 \, \Omega\)

Therefore, the impedance of the circuit is 500 Ω.