ElectroMagnetic InductionHard

Question

In the circuit shown, when dc of 250 V is applied, one ampere current passes through the circuit and when an alternating voltage source of 250 V and 2250 rad/s is applied, current of 1.25 A flows. If maximum current flows through the circuit at a frequency of 4500 rad/s, then the value of L and C will be respectively -
                                       

Options

A.0.295 H, 10-4 F
B.0.0752 H, 0.01 F
C.0.0392 H, 0.005 F
D.0.0494 H, 10-6 F

Solution


In DC (act  as open circuit)
So R = = 250Ω
At resonance ωL =
= 4500L .....(i)
V = VC - VL = VR
I2= V ⇒ I1R = V
Here I = ⇒ I2 =
⇒
- ωL = × 250 here ω = 2250 rad/sec
So, × 250 + 2250L............(ii)
From (i) & (ii) :
2 =
9000L = × 250 + 2250L
L = = 0.0494 H
C = = 10-6 F

Create a free account to view solution

View Solution Free
Topic: ElectroMagnetic Induction·Practice all ElectroMagnetic Induction questions

More ElectroMagnetic Induction Questions

The core of any transformer is laminated so as to...A conducting loop is placed in uniform magnetic field with its plane perpendicular to the field. An emf is induced in th...An induction coil stores 32 joules of magnetic energy and dissipates energy as heat at the rate of 320 watts. when a cur...Which of the following materials is repelled by an external magnetic field :-...A flexible wire bent in the form of a circle is placed in a uniform magnetic field perpendicular to the plane of the coi...