A moving iron voltmeter that has a resistance of 4000 Ω reads the DC supply voltage correctly as 200 V. When connected to measure the AC supply of 200 V, it draws a current of 0.04 A. What is the percentage error in the reading of AC measurement?

Concept

The impedance for the series RL circuit is given by:

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

where, Z = Impedance

R = Resistance

X_L = Reactance

Calculation

Given, R = 4000Ω

V_DC = 200 V

V_AC = 200 V

I_AC = 0.04A

AC impedance, \(Z={V_{AC}\over I_{AC}}\)

\(Z={200\over 0.04}=5000\space \Omega\)

Since the impedance is higher than the DC resistance, it suggests the presence of an inductive reactance X_L.

\(5000=\sqrt{(4000)^2+X_L^2}\)

XL = 3000 Ω

A moving iron voltmeter measures the RMS voltage but is calibrated based on the assumption of pure resistance. The actual voltage it registers is:

V_read = I_AC × R

Vread = 0.04 × 4000 

Vread = 160 V

% error = \({V_{true}-V_{read}\over V_{true}}\times 100\)

% error = \({200-160\over 200}\times 100\)

% error = 20%