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Current Question (ID: 16726)

Question:
$\text{The current in } 8 \, \Omega \text{ resistance is (in the figure below):}$
Options:
  • 1. $0.69 \, \text{A}$
  • 2. $0.92 \, \text{A}$
  • 3. $1.30 \, \text{A}$
  • 4. $1.6 \, \text{A}$
Solution:
$\text{Method - I}$ $-8(I_1 + I_2) - 4I_1 + 8 = 0 \quad \ldots (i)$ $-8(I_1 + I_2) - 6I_2 + 6 = 0 \quad \ldots (ii)$ $\text{Solving eq}^n, (i) \text{ and } (ii), \text{ we get,}$ $I_1 = \frac{8}{13}, \; I_2 = \frac{1}{13}$ $\text{Current in } 8\Omega = I_1 + I_2 = 0.69 \, \text{A}$ $\text{Method II}$ $\text{Given circuit can be reduced to}$ $E_{\text{net}} = \frac{8}{4} + \frac{6}{6} + \frac{4}{6} + \frac{1}{6} = 7.2 \, \text{volt}$ $\frac{1}{R_{\text{net}}} = \frac{1}{4} + \frac{1}{6} + \frac{10}{24}$ $R_{\text{net}} = 2.4 \, \Omega$ $I = \frac{7.2}{10.4} = 0.69 \, \text{A}$

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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
}