Question

A moving iron voltmeter calibrated for 50 Hz supply and having an inductance of the coil as 0.80 H draws 0.1 A when connected to 500 V AC. What will be its reading when connected to 300 V DC supply?

• A. 300.4 V
• B. 150 V
• C. 175.8 V
• D. 325.6 V

Answer 1

The correct option is A 300.4 V

At 50 Hz supply, impedance of the instrument circuits,

$Z=\frac{V_{AC}}{I_{AC}}=\frac{500}{0.1}=5000\Omega$

$R=\sqrt{Z^2-X_L^2}$

$\sqrt{(5000{)}^2-(2\pi \times 50\times 0.8{)}^2}=4993.7\Omega$

When connected to 300 V DC,

Instrument current,

$I_{DC}=\frac{V_{DC}}{R}=\frac{300}{4993.7}=0.06008A$

Reading of instrument when connected to 300 V DC,

$=\frac{V_{DC}}{I_{AC}}\times I_{DC}=\frac{500}{0.1}\times 0.06008=300.4V$