A moving-iron voltmeter has coil resistance 500 Ω and inductance 1 H. It reads 250 V on 250 V DC. With an added 2000 Ω series resistor, what will it read on 250 V, 50 Hz AC? (Assume the scale is effectively linear over this range.)

Introduction / Context:
Moving-iron (MI) voltmeters are commonly used on AC and DC. Their coil has resistance and inductance, so at AC the impedance magnitude is greater than the DC resistance, leading to a slightly smaller current for the same applied RMS voltage. If the instrument scale is set by a DC calibration point, the AC reading can shift because the deflection depends on current (approximately linear in this operating region per the problem statement).

Given Data / Assumptions:

• Coil: R_coil = 500 Ω, L = 1 H.
• Series resistor: R_s = 2000 Ω.
• DC calibration: 250 V produces full-scale indication (i.e., I_DC = 250 / (R_coil + R_s) = 250 / 2500 = 0.1 A).
• AC test: 250 V, f = 50 Hz; assume effective linearity of indication with coil current near this point.

Concept / Approach:
For AC, total impedance Z = √(R_total^2 + (ωL)^2), where R_total = R_coil + R_s and ω = 2πf. Compute AC current I_AC = V / |Z|. With the problem's linear-scale assumption, indicated voltage ≈ 250 * (I_AC / I_DC).

Step-by-Step Solution:

Verification / Alternative check:

Why Other Options Are Wrong:

Common Pitfalls:

Final Answer: