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Current Question (ID: 17303)

Question:

$\text{It is found that the current through the } LCR \text{ series circuit is at its maximum.}$ $\text{If } V_R, V_C \text{ and } V_L \text{ are potential differences across resistance, capacitor, and}$ $\text{inductor respectively, then which of the following is correct?}$

Options:

1. $V_R = V_L > V_C$
2. $V_R \neq V_L = V_C$
3. $V_R = V_L \neq V_C$
4. $V_R = V_C \neq V_L$

Solution:

$\text{In LCR series circuit,}$ $\text{Impedance, } Z = \sqrt{R^2 + (X_L - X_C)^2}$ $I = \frac{E_0 \sin(\omega t \pm \phi)}{Z}$ $\text{Current will be maximum when } Z \text{ will be minimum}$ $i.e; \ X_L = X_C$ $\text{or } V_L = V_C$ $\text{Hence, for maximum current,}$ $V_R \neq V_L = V_C$

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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
}