Question

The potential difference measured across a coil is 4.5 V, when it carries a direct current of 9 A. The same coil when carries an alternating current of 9 A at 25 Hz, the potential difference is 24 V. Find the current, the power and the power factor when it is supplied by 50 V, 50 Hz supply.

Answer 1

Let R be the d.c. resistance and L be the inductance of the coil.

∴ R = V/I = 4.5/9 = 0.5 Ω;

With a.c. current of 25 Hz, Z = V/I = 24/9 = 2.66 Ω.

X_L = √(Z² - R²) = √(2.66² - 0.5²) = 2.62Ω

Now X_L = 2 π × 25 × L ; L = 0.0167 Ω

At 50 Hz

XL = 2.62 × 2 = 5.24 Ω ; Z = √(0.5² + 5.24²) = 5.26Ω

Current I = 50/5.26 = 9.5 A ; Power = I²R = 9.5² × 0.5 = 45W.