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Current Question (ID: 17296)

Question:

$\text{A series AC circuit has a resistance of } 4 \, \Omega \text{ and an inductor of reactance } 3 \, \Omega. \text{ The impedance of the circuit is } Z_1. \text{ Now when a capacitor of reactance } 6 \, \Omega \text{ is connected in series with the above combination, the impedance becomes } Z_2. \text{ Then } \frac{Z_1}{Z_2} \text{ will be:}$

Options:

1. $1 : 1$
2. $5 : 4$
3. $4 : 5$
4. $2 : 1$

Solution:

$\text{Hint: Recall series LCR circuit.}$ $\text{Step 1: Calculate the net impedance of the circuit for Case 1.}$ $Z_1 = \sqrt{R_L^2 + R^2} = 5 \, \Omega$ $\text{Step 2: Calculate the net impedance of the circuit for Case 2.}$ $Z_2 = \sqrt{(R_C - R_L)^2 + R^2} = 5 \, \Omega$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}