Question

A moving-iron voltmeter reads correctly on DC. If connected to a 50 Hz sinusoidal AC source whose RMS value equals the DC value, then the reading will be

• A. greater than the DC reading
• B. less than the DC reading
• C. same as the DC reading
• D. zero

Hint:

Moving-iron instruments are RMS-responding meters. If an AC signal has the same RMS value as a DC signal, a moving-iron meter will show the exact same reading for both.

Answer 1

The Correct Option is C

Concept: Moving-Iron (MI) instruments operate based on the mechanical forces generated between magnetized iron pieces placed within an electromagnetic coil field. The instantaneous deflecting torque developed in any moving-iron instrument is given by the expression: \[ T_d = \frac{1}{2} I^2 \frac{dL}{d\theta} \] Because the deflecting torque is directly proportional to the square of the operating current (\(I^2\)), the steady deflection angle is determined by the mean square current value. Consequently, Moving-Iron instruments naturally read the Root-Mean-Square (RMS) value of an alternating current waveform.

Step 1: Compare the response under DC and AC conditions. When a moving-iron voltmeter is connected to a DC source of value \(V_{dc}\), its steady-state deflection corresponds directly to the value of \(V_{dc}^2\). When connected to an AC source, it responds to the RMS value of the voltage waveform, \(V_{rms}\).

Step 2: Analyze the given conditions. The problem states that the RMS value of the AC source is exactly equal to the DC value: \[ V_{rms} = V_{dc} \] Since the deflecting torque in both scenarios is driven by the exact same effective RMS value, the mechanical pointer will deflect to the same position on the scale. Therefore, the instrument reading remains the same, matching option (C).