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Current Question (ID: 20595)

Question:

$\text{If an insulator with inductive reactance } X_L = R \text{ is connected in series with resistance } R \text{ across an A.C voltage, the power factor comes out to be } P_1. \text{ Now, if a capacitor with capacitive reactance } X_C = R \text{ is also connected in series with the inductor and resistor in the same circuit, the power factor becomes } P_2. \text{ The ratio } \frac{P_1}{P_2} \text{ is:}$

Options:

1. $\sqrt{2} : 1$
2. $1 : \sqrt{2}$
3. $1 : 1$
4. $1 : 2$

Solution:

$\text{Impedance of circuit:}$ $Z = \sqrt{R^2 + R^2} = \sqrt{2}R$ $P_1 = \cos \phi = \text{Power Factor} = \frac{R}{Z} = \frac{1}{\sqrt{2}}$ $\text{So,}$ $\frac{P_1}{P_2} = \frac{\left(\frac{1}{\sqrt{2}}\right)}{1} = \frac{1}{\sqrt{2}}$

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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
}