When a DC voltage of 200 V is applied to a coil of self-inductance ( (2√3/π ) )H, a current of 1 A flows through it. But by replacing DC source with AC source of 200 V , the current in the coil is reduced to 0.5 A . Then the frequency of AC supply is

Question

When a DC voltage of 200 V is applied to a coil of self-inductance ( (2√3/π ) )H, a current of 1 A flows through it. But by replacing DC source with AC source of 200 V , the current in the coil is reduced to 0.5 A . Then the frequency of AC supply is (A) 100 Hz (B) 75 Hz (C) 60 Hz (D) 30 Hz

Tardigrade Admin · Accepted Answer

Resistance of coil (R)=(200/1)200 Ω Current I=(200/√R2+XL2) Or 0.5=(200/√R2+XL2) Or R2+(2π fL)2=(400)2 Or or( 2π f× (2√3/π ) )2=(400)2-(200)2 =120000 Or 4f√3=200√3 Or f=50 Hz

Q. When a $D C$ voltage of $200 V$ is applied to a coil of self-inductance $(\frac{2 3}{π}) H,$ a current of $1 A$ flows through it. But by replacing $D C$ source with $A C$ source of $200 V$ , the current in the coil is reduced to $0.5 A$ . Then the frequency of $A C$ supply is

100 Hz

75 Hz

60 Hz

30 Hz

50 Hz

Resistance of coil $(R) = \frac{200}{1} 200 Ω$
Current $I = \frac{200}{R ^{2} + X _{L}^{2}}$
Or $0.5 = \frac{200}{R ^{2} + X _{L}^{2}}$
Or $R^{2} + (2 π f L)^{2} = (400)^{2}$
Or $or (2 π f \times \frac{2 3}{π})^{2} = (400)^{2} - (200)^{2}$
$= 120000$
Or $4 f 3 = 2003$
Or $f = 50 Hz$

Q. When a DC voltage of 200V is applied to a coil of self-inductance (π23​​)H, a current of 1A flows through it. But by replacing DC source with AC source of 200V , the current in the coil is reduced to 0.5A . Then the frequency of AC supply is