English

When an Inductor is Connected to a 200 V Dc Voltage, a Current at 1a Flows Through It. When the Same Inductor is Connected to a 200 V, 50 Hz Ac Source, Only 0.5 a Current Flows. Explain, Why?

Advertisements
Advertisements

Question

Answer the following question.
When an inductor is connected to a 200 V dc voltage, a current at 1A flows through it. When the same inductor is connected to a 200 V, 50 Hz ac source, only 0.5 A current flows. Explain, why? Also, calculate the self-inductance of the inductor. 

Answer in Brief
Advertisements

Solution

When the inductor is connected across the 200 V DC circuit 1 A current flows because in this case the inductor simply acts as a resistor and there is no inductive reactance. Whereas, when we connect the same inductor across 200 V AC, due to inductive reactance the overall impedance is changed and hence the value of current also changes.  

When the inductor is connected across 200 V DC  

The resistance of the coil 

`R = 200/1 = 200 Ω`

Let the self-inductance of the inductor be L,
When the inductor is connected across 200 V A.C.

Net impedance, Z = `sqrt((2pifL)^2 + R^2) = 200/0.5 = 400 Omega`

⇒ `(2pi xx 50 xx L)^2 + 200^2 = 400^2`

⇒ `(100piL)^2 = 160000 - 40000`

⇒ `10000pi^2L^2 = 120000`

⇒ `L^2 = 12/pi^2`

`therefore L = sqrt(12)/pi H`

shaalaa.com
  Is there an error in this question or solution?
2018-2019 (March) 55/1/3

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

Define the coefficient of self-induction.


Derive an expression for self inductance of a long solenoid of length l, cross-sectional area A having N number of turns.


An average emf of 20 V is induced in an inductor when the current in it is changed from 2.5 A in one direction to the same value in the opposite direction in 0.1 s. Find the self-inductance of the inductor.


A magnetic flux of 8 × 10−4 weber is linked with each turn of a 200-turn coil when there is an electric current of 4 A in it. Calculate the self-inductance of the coil.


Consider a small cube of volume 1 mm3 at the centre of a circular loop of radius 10 cm carrying a current of 4 A. Find the magnetic energy stored inside the cube.


When a coil is connected to a D.C. source of e.m.f. 12 volt, then a current of 4 ampere flows in it. If the same coil is connected to a 12 volt, 25 cycle/s A.C sources, then the current flowing in it is 2.4 A. The self-inductance of the coil will be ______


The magnetic potential energy stored in a certain inductor is 15 mJ, when the current in the inductor is 40 mA. This inductor is of inductance ____________.


When the current in a coil changes from 2 amp. to 4 amp. in 0.05 sec., an e.m.f. of 8 volt is induced in the coil. The coefficient of self inductance of the coil is ______.


A current of 1A flows through a coil when it is connected across a DC battery of 100V. If the DC battery is replaced by an AC source of 100 V and angular frequency of 100 rad s-1, the current reduces to 0.5 A. Find

  1. the impedance of the circuit.
  2. self-inductance of coil.
  3. Phase difference between the voltage and the current.

The current in a coil changes from 50A to 10A in 0.1 second. The self inductance of the coil is 20H. The induced e.m.f. in the coil is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×