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Current Question (ID: 20537)

Question:

$\text{For the LCR circuit, shown here, the current is observed to lead the applied voltage. An additional capacitor } C', \text{ when joined with the capacitor } C \text{ present in the circuit, makes the power factor of the circuit unity. The capacitor } C', \text{ must have been connected in:}$ $V = V_0 \sin \omega t$

Options:

1. $\text{series with } C \text{ and has a magnitude } \frac{C}{\omega^2 LC - 1}$
2. $\text{series with } C \text{ and has a magnitude } \frac{1 - \omega^2 LC}{\omega^2 L}$
3. $\text{parallel with } C \text{ and has a magnitude } \frac{1 - \omega^2 LC}{\omega^2 L}$
4. $\text{parallel with } C \text{ and has a magnitude } \frac{C}{\omega^2 LC - 1}$

Solution:

$\text{Hint: } \cos \phi = \frac{R}{\sqrt{R^2 + (X_L - X_C)^2}}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}