Advertisements
Advertisements
Question
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
- the impedance of the circuit.
- self-inductance of coil.
- Phase difference between the voltage and the current.
Advertisements
Solution
Given: IDC = 1A, V = 100V, ω = 100 rad/sec.
IDC = `V/R ⇒ 100/R = 1`
R = 100 Ω
The a.c. flowing,
IAC = 0.5 A
- Impedance: Z = `sqrt(R^2 + (omegaL)^2`
when AC is applied.
Z = `V/I = 100/0.5 = 200 Omega` - Self-inductance of the coil:
We know `Z^2 = R^2 + X_L^2`
`X_L^2 = Z^2 - R^2`
= `(200)^2 - (100)^2`
= 40000 - 10000
XL = 173.21 Ω
XL = ωL
⇒ L = `X_L/omega = 173.21/100`
L = 1.73 H - The phase difference between V and I, (cosΦ):
`cosphi = R/L = 100/1.73`
`cosphi = 57.73`
`phi = cos^-1(57.73)`
APPEARS IN
RELATED QUESTIONS
Write the SI unit and dimention of of co-efficient of self induction
The currents flowing in the two coils of self-inductance L1 = 16 mH and L2 = 12 mH are increasing at the same rate. If the power supplied to the two coil is equal, find the ratio of the currents ?
A constant current i is maintained in a solenoid. Which of the following quantities will increase if an iron rod is inserted in the solenoid along its axis?
(a) magnetic field at the centre
(b) magnetic flux linked with the solenoid
(c) self-inductance of the solenoid
(d) rate of Joule heating.
An inductor of inductance 5.0 H, having a negligible resistance, is connected in series with a 100 Ω resistor and a battery of emf 2.0 V. Find the potential difference across the resistor 20 ms after the circuit is switched on.
Explain why the inductance of two coils connected in parallel is less than the inductance of either coil.
Two coils of wire A and B are placed mutually perpendicular as shown. When current is changed in any one coil.
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 ______.
The coefficient of self inductance of a solenoid is 0.18 mH. If a core of soft iron of relative permeability 900 is inserted, then the coefficient of self inductance will become nearly ______.
Calculate the self-inductance of a coil using the following data obtained when an AC source of frequency `(200/pi)` Hz and a DC source are applied across the coil.
| AC Source | ||
| S.No. | V (volts) | I (A) |
| 1 | 3.0 | 0.5 |
| 2 | 6.0 | 1.0 |
| 3 | 9.0 | 1.5 |
| DC Source | ||
| S.No. | V (volts) | I (A) |
| 1 | 4.0 | 1.0 |
| 2 | 6.0 | 1.5 |
| 3 | 8.0 | 2.0 |
In a coil, the current changes form −2 A to +2 A in 0.2 s and induces an emf of 0.1 V. The selfinductance of the coil is ______.
