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Current Question (ID: 17324)

Question:

$\text{In the circuit shown, the value of RMS current is 11 A. The potential difference across the inductor is:}$

Options:

1. $220 \text{ V}$
2. $\text{zero}$
3. $300 \text{ V}$
4. $200 \text{ V}$

Solution:

$\text{Hint: Recall series LCR circuit.}$ $\text{Step 1: Calculate the voltage across resistance.}$ $\text{Step 2: Calculate the voltage across the inductor.}$ $V_N = \sqrt{V_R^2 + (V_L - V_C)^2}$ $220 = \sqrt{(220)^2 + (V_L - V_C)^2}$ $V_L = V_C = 200 \text{ V}$

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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
}