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Current Question (ID: 17321)

Question:
$\text{In an } LCR \text{ series AC circuit, the voltage across each of the components } L, C \text{ and } R \text{ is } 50 \text{ V. The voltage across the } LR \text{ combination will be:}$
Options:
  • 1. $50 \text{ V}$
  • 2. $50\sqrt{2} \text{ V}$
  • 3. $100 \text{ V}$
  • 4. $0 \text{ V}$
Solution:
$\text{Hint: } V = \sqrt{V_L^2 + V_R^2}.$ $\text{Step 1: Draw phasor diagram}$ $\text{Step 2: Find resultant voltage of inductor \& resistor}$ $V_N = \sqrt{V_L^2 + V_R^2} = 50\sqrt{2}V$

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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
}