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Current Question (ID: 16547)

Question:

$\text{Three uncharged capacitors of capacities } C_1, C_2 \text{ and } C_3 \text{ are connected to one another as shown in the figure.}$ $\text{If points A, B, and D, are at potential } V_1, V_2 \text{ and } V_3 \text{ then the potential at O will be:}$

Options:

1. $\frac{V_1C_1 + V_2C_2 + V_3C_3}{C_1 + C_2 + C_3}$
2. $\frac{V_1 + V_2 + V_3}{C_1 + C_2 + C_3}$
3. $\frac{V_1(V_2 + V_3)}{C_1(C_2 + C_3)}$
4. $\frac{V_1V_2V_3}{C_1C_2C_3}$

Solution:

$\text{Hint: Use } Q = CV$ $\text{Step: Find the potential at the point O.}$ $\text{Let the potential at point O is } V_0$ $\text{By using Kirchhoff's law we get;}$ $\Rightarrow C_1(V_0 - V_1) + C_2(V_0 - V_2) + C_3(V_0 - V_3) = 0$ $\text{Solving the above equation we get;}$ $\Rightarrow V_0 = \frac{V_1C_1 + V_2C_2 + V_3C_3}{C_1 + C_2 + C_3}$ $\text{Hence, option (1) is the correct answer.}$

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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
}