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Current Question (ID: 20538)

Question:

$\text{In an ac circuit, the instantaneous e.m.f. and current are given by}$ $e = 100 \sin 30t$ $i = 20 \sin \left(30t - \frac{\pi}{4}\right)$ $\text{In one cycle of ac, the average power consumed by the circuit and the wattless current are, respectively:}$

Options:

1. $50, 10$
2. $\frac{1000}{\sqrt{2}}, 10$
3. $\frac{50}{\sqrt{2}}, 0$
4. $50, 0$

Solution:

$\text{Hint: } P = \frac{e_0 i_0 \cos \phi}{2}$ $e = 100 \sin 30t$ $i = 20 \sin \left(30t - \frac{\pi}{4}\right)$ $P = \frac{e_0 i_0 \cos \phi}{2}$ $= \frac{100 \times 20}{2} \times \cos \frac{\pi}{4}$ $P = \frac{1000}{\sqrt{2}} \text{ watt}$ $\text{Wattless current}$ $I = \frac{(I_0 \sin \phi)}{\sqrt{2}}$ $= \frac{20}{2} = 10 \text{ A}$

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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
}