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Current Question (ID: 20556)

Question:

$\text{Evaluate the maximum current flowing through the capacitor in the given AC circuit.}$ $C = 150 \, \mu\text{F}$ $E = 36 \sin(120 \, \pi t) \, \text{V}$

Options:

1. 0.65 \pi \, \text{A}
2. 0.35 \pi \, \text{A}
3. 0.2 \pi \, \text{A}
4. 0.8 \pi \, \text{A}

Solution:

$\text{Hint: } I_0 = \frac{E_0}{X_C}$ $\text{Step: Find the maximum current flowing through the given capacitor.}$ $\text{The maximum current flowing through the given circuit is given by:}$ $i_0 = \frac{E_0}{X_C} = E_0 \, \omega C \left( X_C = \frac{1}{\omega C} \right)$ $\Rightarrow i_0 = 36 \times 120 \pi \times 150 \times 10^{-6} \, \text{A}$ $\Rightarrow i_0 = 0.65 \pi \, \text{A}$ $\text{Hence, option (1) is the correct answer.}$

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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
}