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Current Question (ID: 20593)

Question:

$\text{A circuit element } X \text{ when connected to an AC supply of peak voltage } 100 \text{ V gives a peak current of } 5 \text{ A which is in phase with the voltage.}$ $\text{A second element } Y \text{ when connected to the same AC supply also gives the same value of peak current which lags behind the voltage by } \frac{\pi}{2}. \text{ If } X \text{ and } Y \text{ are connected in series to the same supply, the RMS value of the current in ampere will be:}$

Options:

1. $\frac{10}{\sqrt{2}}$
2. $\frac{5}{\sqrt{2}}$
3. $5\sqrt{2}$
4. $\frac{5}{2}$

Solution:

$\text{Hint: } X_L = \omega L$

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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
}