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Current Question (ID: 20559)

Question:

$\text{In a series } LCR \text{ circuit, the resistance } R, \text{ inductance } L, \text{ and}$ $\text{capacitance } C \text{ are } 10 \, \Omega, \, 0.1 \, \text{H, and } 2 \, \text{mF, respectively. If the}$ $\text{angular frequency of the AC source is } 100 \, \text{rad/s, the power factor of}$ $\text{the circuit is:}$

Options:

1. $\frac{1}{\sqrt{5}}$
2. $\frac{2}{\sqrt{5}}$
3. $\frac{3}{\sqrt{5}}$
4. $\frac{2}{2\sqrt{5}}$

Solution:

$\text{Hint: } \cos \phi = \frac{R}{Z}$ $\text{Step 1: Find the impedance of the circuit.}$ $\text{The impedance of the inductor is given by:}$ $X_L = \omega L = 0.1 \times 100 = 10 \, \Omega$ $\text{The impedance of the capacitor is given by:}$ $X_C = \frac{1}{\omega C} = \frac{1}{100 \times 2 \times 10^{-3}} = 5 \, \Omega$ $\text{The impedance of the circuit is given by:}$ $Z = \sqrt{R^2 + (X_L - X_C)^2}$ $\Rightarrow Z = \sqrt{(10)^2 + (10 - 5)^2} = \sqrt{100 + 25} = \sqrt{125}$ $\Rightarrow Z = 5\sqrt{5} \, \Omega$ $\text{Step 2: Find the power factor of the circuit.}$ $\text{The power factor of the circuit is given by:}$ $\cos \phi = \frac{R}{Z}$ $\Rightarrow \cos \phi = \frac{10}{5\sqrt{5}} = \frac{2}{\sqrt{5}}$ $\text{Hence, option (2) is the correct answer.}$

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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
}