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प्रश्न
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.
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उत्तर
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)`
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(a) self-inductance
(b) rate of Joule heating if the same current goes through them
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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 |
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