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Current Question (ID: 17327)

Question:

$\text{An ac source of angular frequency } \omega \text{ is fed across a resistor } r \text{ and a capacitor } C \text{ in series. } I \text{ is the current in the circuit. If the frequency of the source is changed to } \frac{\omega}{3} \text{ (but maintaining the same voltage), the current in the circuit is found to be halved. Calculate the ratio of reactance to resistance at the original frequency } \omega.$

Options:

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

Solution:

$\text{(1) At angular frequency } \omega, \text{ the current in } RC \text{ circuit is given by}$ $i_{\text{rms}} = \frac{V_{\text{rms}}}{\sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2}} \quad \text{......(i)}$ $\text{Also } \frac{i_{\text{rms}}}{2} = \frac{V_{\text{rms}}}{\sqrt{R^2 + \frac{9}{\omega^2 C^2}}} \quad \text{......(ii)}$ $\text{From equation (i) and (ii) we get}$ $3R^2 = \frac{5}{\omega^2 C^2} \Rightarrow \frac{1}{\omega C} = R \Rightarrow \sqrt{\frac{3}{5}}$ $\Rightarrow \frac{X_C}{R} = \sqrt{\frac{3}{5}}$

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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
}