In a 1 kHz series RL circuit, the resistor R and the inductive reactance XL are both 1000 Ω. What is the magnitude of the total impedance seen by the source?

Introduction / Context:
This problem tests series RL impedance calculation. In AC circuit analysis, impedance combines resistance and reactance vectorially. Because resistor voltage is in phase with current while inductor voltage leads current by 90 degrees, the total magnitude is found using the Pythagorean relationship, not simple arithmetic addition.

Given Data / Assumptions:

• Frequency f = 1 kHz (value does not affect the result once XL is given).
• R = 1000 Ω (purely resistive).
• XL = 1000 Ω (inductive reactance).
• Series connection of R and L (same current through both).

Concept / Approach:
Impedance of a series RL network is Z = R + jXL. The magnitude is |Z| = sqrt(R^2 + XL^2). The phase angle is theta = arctan(XL / R), though only the magnitude is requested here. Since R and XL are equal, the triangle is isosceles, and |Z| will be R * sqrt(2).

Step-by-Step Solution:
Compute magnitude: |Z| = sqrt(R^2 + XL^2)Insert values: |Z| = sqrt(1000^2 + 1000^2)|Z| = sqrt(1,000,000 + 1,000,000) = sqrt(2,000,000)|Z| ≈ 1414 Ω (since 1000 * sqrt(2) ≈ 1414.21)

Verification / Alternative check:
Because R = XL, the phase angle is arctan(1) = 45 degrees. A 45-degree impedance triangle has hypotenuse R * sqrt(2), confirming the numeric result without a calculator.

Why Other Options Are Wrong:

• 500 Ω and 707 Ω: These correspond to parallel combinations or RMS-like misapplications, not series magnitude.
• 1000 Ω: Ignores the inductive reactance.
• 2000 Ω: Simple arithmetic sum, not vector sum, so incorrect.

Common Pitfalls:

• Adding R and XL directly instead of using the square root of the sum of squares.
• Confusing magnitude with phase; both are important, but only the magnitude is asked.

Final Answer:
1414 Ω