Question

In an L-C-R series circuit, the values of R, $XL$ and $XC$ are 120 Ω, 180 Ω and 130 Ω. What is the impedance of the circuit?

• A. 120 Ω
• B. 130 Ω
• C. 180 Ω
• D. 330 Ω

Hint:

$Z = \sqrtR² + (XL - XC)²}$.

Answer 1

The Correct Option is B

To determine the impedance of an L-C-R series circuit, we need to calculate the net impedance using the formula for impedance in a series circuit, which is given by:

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

where:

• \(R\) is the resistance.
• \(X_L\) is the inductive reactance.
• \(X_C\) is the capacitive reactance.

Given values are:

• \(R = 120 \, \Omega\)
• \(X_L = 180 \, \Omega\)
• \(X_C = 130 \, \Omega\)

First, calculate the net reactance:

\(X = X_L - X_C = 180 - 130 = 50 \, \Omega\)

Substitute the known values into the impedance formula:

\(Z = \sqrt{(120)^2 + (50)^2}\)

Calculate the squares:

\(Z = \sqrt{14400 + 2500}\)

Sum the squares:

\(Z = \sqrt{16900}\)

Find the square root:

\(Z = 130 \, \Omega\)

Therefore, the impedance of the circuit is 130 Ω.

Conclusion: The correct answer is 130 Ω. This matches the given correct answer in the options, confirming that our calculations are accurate.