मराठी

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 - Physics

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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

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

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

  1. Impedance: Z = `sqrt(R^2 + (omegaL)^2`
    when AC is applied.
    Z = `V/I = 100/0.5 = 200 Omega`
  2. 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
  3. 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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