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Current Question (ID: 20136)

Question:

$\text{Consider the combination of 2 capacitors } C_1 \text{ and } C_2, \text{ with } C_2 > C_1,$ $\text{when connected in parallel, the equivalent capacitance is } \frac{15}{4} \text{ times}$ $\text{the equivalent capacitance of the same connected in series.}$ $\text{Calculate the ratio of capacitors, } \frac{C_2}{C_1}.$

Options:

1. $\frac{15}{11}$
2. $\frac{111}{80}$
3. $\frac{29}{15}$
4. $\text{None of the above}$

Solution:

$\text{Recall the series and parallel combination of capacitors.}$ $C_{\text{eq}} = C_1 + C_2$ $C'_{\text{eq}} = \frac{C_1 C_2}{C_1 + C_2}$ $C_1 + C_2 = \frac{15}{4} \left( \frac{C_1 C_2}{C_1 + C_2} \right)$ $4(C_1 + C_2)^2 = 15C_1C_2 = 0$ $4C_1^2 + 4C_2^2 - 7C_1C_2 = 0$ $\text{dividing by } C_1^2$ $4 \left( \frac{C_2}{C_1} \right)^2 - \frac{7C_2}{C_1} + 4 = 0$ $\text{Let } \frac{C_2}{C_1} = x$ $4x^2 - 7x + 4 = 0$ $b^2 - 4ac = 49 - 64 < 0$ $\text{No solution exists}$

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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
}