A 250 V moving iron voltmeter takes a current of 0.05 A when connected to a 250 V DC supply. The coil has an inductance of 1 H. What will be the reading on the meter when connected to a 250 V, 100 Hz supply? Assume impedance at 100 Hz to be 6000 Ω.

Concept

The voltage reading of a moving iron voltmeter corresponds to the true RMS voltage across its resistance:

V_meter = I_AC × R

where, I_AC = AC current

R = DC resistance

Calculation

Given, V_DC = 250 V

I_DC = 0.05 A

L = 1 H

f = 100 Hz

Impedance at 100Hz, Z_AC = 6000 Ω

\(R={V_{DC}\over I_{DC}}\)

\(R={250\over 0.05}=5000\space Ω\)

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

\(X_L=2\pi fL=2\pi (1000)(1)=628.32\space Ω\)

\(Z_{AC}=\sqrt{(5000)^2+(628.32)^2}\)

Z_AC = 5040 Ω 

\(I_{AC}={V_{AC}\over Z_{AC}}\)

\(I_{AC}={250\over 6000}=0.04167A\)

Vmeter = (0.04167) × (5000)

Vmeter = 208.33 V