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Current Question (ID: 17314)

Question:

$\text{For a series } RLC \text{ circuit, } R = X_L = 2X_C. \text{ The impedance of the circuit and phase difference between } V \text{ and } i \text{ will be:}$

Options:

1. $\frac{\sqrt{5}R}{2}, \tan^{-1}(2)$
2. $\frac{\sqrt{5}R}{2}, \tan^{-1}\left(\frac{1}{2}\right)$
3. $\sqrt{5}X_C, \tan^{-1}(2)$
4. $\sqrt{5}R, \tan^{-1}\left(\frac{1}{2}\right)$

Solution:

$X_L = R, \ X_C = \frac{R}{2}$ $\therefore \tan \phi = \frac{X_L - X_C}{R} = \frac{R - \frac{R}{2}}{R} = \frac{1}{2}$ $\Rightarrow \phi = \tan^{-1}\left(\frac{1}{2}\right)$ $\text{Also } Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + \frac{R^2}{4}} = \frac{\sqrt{5}}{2}R$

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}