मराठी

An Alternating Emf of 110 V is Applied to a Circuit Containing a Resistance R of 80 ω and an Inductor L in Series. the Current is Found to Lag Behind the Supply Voltage by an Angle 8 = Tan-1 (3/4). - Physics (Theory)

Advertisements
Advertisements

प्रश्न

An alternating emf of 110 V is applied to a circuit containing a resistance R of 80 Ω and an inductor L in series. The current is found to lag behind the supply voltage by an angle 8 = tan-1 (3/4). Find the :
(i) Inductive reactance
(ii) Impedance of the circuit
(iii) Current flowing in the circuit
(iv) If the inductor has a coefficient of self-inductance of 0.1 H, what is the frequency of the applied emf?

बेरीज

उत्तर

(i) Here, Ev = 110 V, R = 80 Ω, L = 0.1 H
θ = tan-1`(3/4)`

We know that in an L - R circuit
tan θ = `"Lω"/"R" = "X"_"L"/"R"`

`3/4 = "X"_"L"/80`

`"X"_"L" = 80 x 3/4` = 60 Ω

(ii) impedance Z = `sqrt( "X"_L^2 + "R"^2 )`

Z = `sqrt( (60)^2 + (80)^2 )`

 Z = 100 Ω

(ii) Now,
`"I"_"v" = "E"_"v"/"Z" = 110/100 = 1.1 "A"`

(iv) XL = 2πnL

∴ n = `"X"_"L"/(2π"L") = 60/( 2 xx 3.14 xx 0.1 ) = 95.54 "Hz"`.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
2013-2014 (March)

APPEARS IN

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

(a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side an as shown in the figure.

(b) Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, v = 10 m/s.

Calculate the induced emf in the loop at the instant when x = 0.2 m.

Take a = 0.1 m and assume that the loop has a large resistance.


A rod AB moves with a uniform velocity v in a uniform magnetic field as shown in figure.


The flux of magnetic field through a closed conducting loop changes with time according to the equation, Φ = at2 + bt + c. (a) Write the SI units of a, b and c. (b) If the magnitudes of a, b and c are 0.20, 0.40 and 0.60 respectively, find the induced emf at t = 2 s.


Suppose the resistance of the coil in the previous problem is 25Ω. Assume that the coil moves with uniform velocity during its removal and restoration. Find the thermal energy developed in the coil during (a) its removal, (b) its restoration and (c) its motion.


A closed coil having 100 turns is rotated in a uniform magnetic field B = 4.0 × 10−4 T about a diameter which is perpendicular to the field. The angular velocity of rotation is 300 revolutions per minute. The area of the coil is 25 cm2 and its resistance is 4.0 Ω. Find (a) the average emf developed in half a turn from a position where the coil is perpendicular to the magnetic field, (b) the average emf in a full turn and (c) the net charge displaced in part (a).


The current generator Ig' shown in figure, sends a constant current i through the circuit. The wire cd is fixed and ab is made to slide on the smooth, thick rails with a constant velocity v towards right. Each of these wires has resistance r. Find the current through the wire cd.


An inductor-coil of inductance 20 mH having resistance 10 Ω is joined to an ideal battery of emf 5.0 V. Find the rate of change of the induced emf at (a) t = 0,  (b) t = 10 ms and (c) t = 1.0 s.


The emf is induced in a single, isolated coil due to ...A...of flux through the coil by means of varying the current through the same coil. This phenomenon is called ...B... Here, A and B refer to ______.

A conducting square loop of side 'L' and resistance 'R' moves in its plane with the uniform velocity 'v' perpendicular to one of its sides. A magnetic induction 'B' constant in time and space pointing perpendicular and into the plane of the loop exists everywhere as shown in the figure. The current induced in the loop is ______.


A rectangular loop of sides 8 cm and 2 cm with a small cut is stationary in a uniform magnetic field directed normal to the loop. The magnetic field is reduced from its initial value of 0.3 T at the rate of 0.02 T s-1 If the cut is joined and loop has a resistance of 1.6 Ω, then how much power is dissipated by the loop as heat?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×