Import Question JSON

Current Question (ID: 17299)

Question:

$\text{Choose the correct statements regarding a series LCR circuit.}$ $1.\ \text{The voltage across the capacitor lags behind the current.}$ $2.\ \text{The voltage across the inductor leads the current.}$ $3.\ \text{The voltage across the resistance } R \text{ is in phase with the current.}$ $4.\ \text{All of the above.}$

Options:

1. $\text{The voltage across the capacitor lags behind the current.}$
2. $\text{The voltage across the inductor leads the current.}$
3. $\text{The voltage across the resistance } R \text{ is in phase with the current.}$
4. $\text{All of the above.}$

Solution:

$\text{Recall the phasor diagram.}$ $\text{The phasor diagram for a series } L, C, R \text{ circuit is shown in the figure below;} \ V_L, V_C, V_R, I$ $\text{In a capacitor, the voltage lags behind the current by } 90^\circ. \text{ This is because, in a capacitor, the current leads to the voltage, as the capacitor takes time to charge.}$ $\text{In an inductor, the voltage leads the current by } 90^\circ. \text{ This is because the inductor resists changes in current, creating a phase difference where the voltage leads the current.}$ $\text{In a resistor, the voltage and current are in phase. There is no phase difference because the resistor does not store energy like capacitors or inductors.}$ $\text{Therefore, all the statements are correct.}$ $\text{Hence, option (4) is the correct answer.}$

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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
}