Q138 / 150 · MHT CET 2024 (03 May Shift 1) · Physics

An inductance coil has a resistance of $80 \Omega$. When on AC signal of frequency 480 Hz is applied to the coil, the voltage leads the current by $45^{\circ}$. The inductance of the coil in henry is $\left[\sin 45^{\circ}=\cos 45^{\circ}=1 / \sqrt{2}\right]$

  • A
    $\frac{1}{24 \pi}$
  • B
    $\frac{\pi}{20}$
  • C
    $\frac{\pi}{40}$
  • Correct answer
    D
    $\frac{1}{12 \pi}$

Solution

$\begin{array}{ll} & \tan \phi=\frac{X_L}{R} \\ & \tan 45=1=\frac{2 \pi \mathrm{fL}}{\mathrm{R}} \\ \therefore \quad & L=\frac{R}{2 \pi \mathrm{f}}=\frac{80}{2 \pi \times 480}=\frac{1}{12 \pi}\end{array}$

Hence, the correct answer is option: 'D'.

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