Question

The impedance of L-R circuit is

• A. \(R^2 + \omega^2L^2\)
• B. \(R + \omega L\)
• C. \(\sqrt{R + \omega L}\)
• D. \(\sqrt{R^2 + \omega^2L^2}\)

Hint:

For any series RLC circuit, the impedance is \(Z = \sqrt{R^2 + (X_L - X_C)^2}\). For an L-R circuit, there is no capacitor, so \(X_C = 0\), which simplifies the formula to \(Z = \sqrt{R^2 + X_L^2}\). For an R-C circuit, \(X_L = 0\), so \(Z = \sqrt{R^2 + (-X_C)^2} = \sqrt{R^2 + X_C^2}\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Impedance (\(Z\)) is the total opposition that a circuit presents to the flow of alternating current. In a series L-R circuit, it arises from two sources: the resistance (\(R\)) of the resistor and the inductive reactance (\(X_L\)) of the inductor. These two quantities do not add arithmetically because the voltage across the resistor is in phase with the current, while the voltage across the inductor leads the current by 90 degrees.
Step 2: Key Formula or Approach:
The resistance is \(R\).
The inductive reactance is given by \(X_L = \omega L\), where \(\omega\) is the angular frequency of the AC supply.
Since \(R\) and \(X_L\) are out of phase by 90°, they are combined as vectors (or phasors) using the Pythagorean theorem. This can be visualized using an impedance triangle where R is the adjacent side, \(X_L\) is the opposite side, and Z is the hypotenuse. \[ Z = \sqrt{R^2 + X_L^2} \] Step 3: Detailed Explanation:
Substitute the expression for inductive reactance (\(X_L = \omega L\)) into the general formula for impedance: \[ Z = \sqrt{R^2 + (\omega L)^2} \] This simplifies to: \[ Z = \sqrt{R^2 + \omega^2 L^2} \] Step 4: Final Answer:
Comparing the derived expression with the given options, we see that option (D) correctly represents the impedance of a series L-R circuit.