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Current Question (ID: 17326)

Question:
$\text{An alternating voltage source is connected to a series } RC \text{ circuit. Consider two situations:}$ $1. \text{When the capacitor is air-filled.}$ $2. \text{When the capacitor is mica filled.}$ $\text{If the current through the resistor is } I \text{ and the voltage across the capacitor is } V, \text{ then:}$
Options:
  • 1. $V_a \leq V_b$
  • 2. $V_a > V_b$
  • 3. $i_a > i_b$
  • 4. $V_a = V_b$
Solution:
$V = V_0 \sin \omega t$ $X_c = \frac{1}{2 \pi f C}$ $\text{So, current in circuit, } I = \frac{V}{Z} = \frac{V}{\sqrt{R^2 + \left(\frac{1}{2 \pi f C}\right)^2}}$ $I = \frac{2 \pi f C}{\sqrt{4 \pi^2 f^2 C^2 R^2 + 1}} \times V$ $\text{The voltage drops across the capacitor, } V_c = I \times X_c$ $= \frac{2 \pi f C}{\sqrt{4 \pi^2 f^2 C^2 R^2 + 1}} \times \frac{1}{2 \pi f C} \times V$ $i.e., \ V_c = \frac{V}{\sqrt{(4 \pi^2 f^2 C^2 R^2 + 1)}}$ $\text{When mica is introduced, capacitance increases. Hence, the voltage across the capacitor decreases.}$

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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
}