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Current Question (ID: 17370)

Question:
$\text{An inductor of } 20 \text{ mH, a capacitor of } 50 \ \mu\text{F, and a resistor of } 40 \, \Omega \text{ are connected in series across a source of emf } V = 10 \sin 340t. \text{ What will be the power loss in the AC circuit?}$
Options:
  • 1. $0.67 \text{ W}$
  • 2. $0.78 \text{ W}$
  • 3. $0.89 \text{ W}$
  • 4. $0.46 \text{ W}$
Solution:
$\text{Given:}$ $L = 20 \text{mH}, \ C = 50 \mu\text{F and } V = 10 \sin 340t$ $\text{Power loss in AC circuit, } P_{\text{av}} = I_v^2 R = \left( \frac{E_v}{Z} \right)^2 R$ $P_{\text{av}} = \left( \frac{10}{\sqrt{2}} \right)^2 \times \frac{1}{\sqrt{(40)^2 + \left(340 \times 20 \times 10^{-3} - \frac{1}{340 \times 50 \times 10^{-6}}\right)^2}} \times 40$ $P_{\text{av}} = 0.46 \text{W}$

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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
}