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

A coil of wire having finite inductance and resistance has conducting ring placed co-axially with in it. The coil is con...The loop shown moves with a constant velocity ′v′ in a uniform magnetic field of magnitude ′B′ d...A choke coil consists of 150 turns wound on a high permeability core. If the relative permeability is 3540, permeability...Two inductor coils of self inductance 3H and 6H respectively are connected with a resistance 10Ω and a battery 10 V...When a current changes from 2A to 4A in 0.05 sec. in a coil, induced emf is 8 V. The self inductance of coil is :-...