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Current Question (ID: 20230)

Question:

$\text{A current of } 6 \text{ A enters at the corner } P \text{ of an equilateral triangle } PQR, \text{ with each side having a wire of resistance } 2 \, \Omega. \text{ The current exits at corner } R. \text{ The current } i_1 \text{ in ampere is:}$

Options:

1. $2$
2. $3$
3. $6$
4. $9$

Solution:

$\text{Hint: Apply Kirchhoff's junction law.}$ $\text{Step: Find the current } i_1 \text{ in the circuit.}$ $\text{According to Kirchhoff's junction rule, } i_1 + i_2 = 6 \, \text{A}$ $\text{By the current division rule, the current through } i_1 \text{ is given by:}$ $i_1 = \frac{R_{PQ} \times I}{R_{PR} + R_{PQ}} \text{ where, } R_{PQ} \text{ is the resistance across the wire } PQ \text{ and } R_{PR} \text{ is the resistance across the wire } PR.$ $\Rightarrow i_1 = \frac{2 \times 6}{2 + 4} = \frac{12}{6} = 2 \, \text{A}$ $\text{Hence, option (1) is the correct answer.}$

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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
}