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Current Question (ID: 17307)

Question:
$\text{A sinusoidal supply of frequency } 10 \text{ Hz and rms voltage of } 12 \text{ V is connected to a } 2.1 \, \mu\text{F capacitor. What is the rms value of current?}$
Options:
  • 1. $5.5 \, \text{mA}$
  • 2. $20 \, \text{mA}$
  • 3. $26 \, \text{mA}$
  • 4. $1.6 \, \text{mA}$
Solution:
$\text{Hint: } I_{\text{rms}} = \frac{V_{\text{rms}}}{X_C}$ $\text{Step 1: Find the capacitive reactance}$ $X_C = \frac{1}{C \omega}$ $\omega = 2 \pi f$ $\text{Step 2: Find the r.m.s. current}$ $I_{\text{rms}} = \frac{V_{\text{rms}}}{X_C} = 12 \times 2.1 \times 10^{-6} \times 2 \pi \times 10 \approx 1.6 \, \text{mA}$

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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
}