Energy Stored in an Inductor
Trending Questions
Q. The current in a coil of self-inductance 2.0 H is increasing according to equation I=2sint2 A. Find the amount of energy spent during the period when the current changes from 0 to 2 A.
- 3 J
- 4 J
- 2 J
- None of these
Q. A short-circuited coil is placed in a time-varying magnetic field. Electrical power is dissipated due to the current induced in the coil. If the number of turns were to be quadrupled and the wire radius halved, the electrical power dissipated would be
- Halved
- The same
- Doubled
- Quadrupled
Q. A coil of resistance 20 Ω and inductance 5 H has been connected to a 100 V battery The energy stored in the coil after a long time is
- 31.25 J
- 125 J
- 62.5 J
- 250 J
Q. Two different coils have self-inductance L1=8mH, L2=2mH. The current in one coil is increased at a constant rate. The current in the second coil is also increased at the same rate. At a certain instant of time, the power given to the two coils is the same. At that time the current, the induced voltage and the energy stored in the first coil are i1, V1 and W1 respectively. Corresponding values for the second coil at the same instant are i2, V2 and W2 respectively. Then
- i1i2=14
- V2V1=12
- i1i2=4
- W2W1=2
Q. Shown in the figure is a circular loop of radius r and resistance R. A variable magnetic field of induction B = B0e−t is established inside the coil. If the key (K) is closed, the electrical power developed right after closing the switch is equal to
Q.
A coil of inductance L is carrying a steady current i. What is the nature of its stored energy?
Magnetic
Electrical
Heat
Both magnetic and electrical
Q. An electron is shot into one end of a solenoid at the speed of 800 m/s making an angle 30° with the axis of solenoid. Current in solenoid is 4 A and there are total 8000 turns over solenoid. If the total number of revolutions made by electron within solenoid before emerging from another end of solenoid is n×106, then the value of n is
(em=√3×1011 C/kg (write two digits after the decimal point.)]
(em=√3×1011 C/kg (write two digits after the decimal point.)]