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Current Question (ID: 17365)

Question:
$\text{A resistance } R \text{ draws power } P \text{ when connected to an AC source. If an inductance is now placed in series with the resistance, such that the impedance of the circuit becomes } Z \text{ the power drawn will be:}$
Options:
  • 1. $P \left( \frac{R}{Z} \right)^2$
  • 2. $P \sqrt{\frac{R}{Z}}$
  • 3. $P \left( \frac{R}{Z} \right)$
  • 4. $P$
Solution:
$P = V_{\text{rms}} \times I_{\text{rms}} = V_{\text{rms}} \times \frac{V_{\text{rms}}}{R}$ $\Rightarrow V_{\text{rms}}^2 = PR$ $\text{When an Inductor is connected in series with resistor then power drawn—}$ $P' = V_{\text{rms}} I_{\text{rms}} \cos \phi$ $\text{Where } \phi \text{ is the phase difference}$ $P' = \frac{V_{\text{rms}}^2}{R} \times \frac{R^2}{Z^2}$ $\Rightarrow P' = P \left( \frac{R}{Z} \right)^2$

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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
}