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Current Question (ID: 20598)

Question:

$\text{In a series } LCR \text{ circuit connected across } (220 \text{ V, } 50 \text{ Hz) ac supply. If the inductive reactance of the circuit is } 79.6 \, \Omega. \text{ If the power delivered in the circuit is maximum, then the capacitance of the circuit is:}$

Options:

1. $40 \, \mu F$
2. $30 \, \mu F$
3. $20 \, \mu F$
4. $50 \, \mu F$

Solution:

$\text{For maximum power, LCR should be in a resonance condition-}$ $X_L = X_C$ $\Rightarrow \; 79.6 = \frac{1}{\omega C} = \frac{1}{2 \pi f C} = \frac{1}{2 \pi \times 50 \times C}$ $\Rightarrow \; C = \frac{1}{79.6 \times 100 \pi} = 40 \times 10^{-6} \text{ F} = 40 \, \mu F$

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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
}