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Current Question (ID: 17350)

Question:
$\text{In an AC circuit, the current is given by;} \; i = 5 \sin(100t - \frac{\pi}{2}) \; \text{and the AC potential is} \; V = 200 \sin(100t) \; \text{V. The power consumption is:}$
Options:
  • 1. $20 \; \text{W}$
  • 2. $40 \; \text{W}$
  • 3. $1000 \; \text{W}$
  • 4. $\text{zero}$
Solution:
$\text{Hint:} \; P_{\text{avg}} = V_0 I_0 \cos \phi$ $\text{Step: Find the power consumption in the AC circuit.}$ $\text{The power consumption in the AC circuit is given by;} \; P_{\text{avg}} = V_0 I_0 \cos \phi$ $\text{The phase difference;} \; \phi = \frac{\pi}{2}$ $\text{Substitute the values we get;} \; P_{\text{avg}} = V_0 I_0 \times \frac{\pi}{2}$ $\Rightarrow \; P_{\text{avg}} = 0$ $\text{Hence, option (4) is the correct answer.}$

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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
}