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Current Question (ID: 20205)

Question:
$\text{In the circuit shown, three } 1 \, \Omega \text{ resistors are connected vertically between the upper and lower conductors. Each section of both the upper and lower conductors contains an ideal } 2 \, \text{V source, all oriented in the same direction. The left ends of the two conductors are directly connected.}$ $\text{What is the current through each of the } 1 \, \Omega \text{ resistor?}$
Options:
  • 1. $1 \, \text{A}$
  • 2. $0.25 \, \text{A}$
  • 3. $0.5 \, \text{A}$
  • 4. $0 \, \text{A}$
Solution:
$\text{Hint: Potential difference across each resistor is zero so current in each resistor also zero.}$ $\text{Step 1: Find the potential at each nodes.}$ $\text{Step 2: Find the current in each resistor.}$ $\text{Since the potential difference across each resistor is zero, the current through each resistor is also zero.}$ $\text{Hence, option (4) is the correct answer.}$

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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
}