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Current Question (ID: 16660)

Question:

$\text{If power dissipated in the } 9 \, \Omega \text{ resistor in the circuit shown is } 36 \, \text{W, the potential difference across the } 2 \, \Omega \text{ resistor will be:}$

Options:

1. $8 \, \text{V}$
2. $10 \, \text{V}$
3. $2 \, \text{V}$
4. $4 \, \text{V}$

Solution:

$\text{(2) Hint: The current in the parallel branch will be in the inverse ratio of resistances.}$ $\text{Step 1: Find the current in } 9 \, \Omega \text{ and } 6 \, \Omega \text{ resistance.}$ $\text{Power is given by, } P = i^2 R$ $\therefore \text{ Current, } i = \sqrt{\frac{P}{R}}$ $\text{For resistance of } 9 \, \Omega,$ $i_1 = \sqrt{\frac{36}{9}} = \sqrt{4} = 2 \, \text{A}$ $\text{As } 9 \, \Omega \text{ and } 6 \, \Omega \text{ are in parallel:}$ $i_2 = \frac{i_1 \times R_1}{R_2} = \frac{2 \times 9}{6} = 3 \, \text{A}$ $\text{Step 2: Find the potential drop in } 2 \, \Omega \text{ resistance.}$ $\text{Here } (9 \Omega \parallel 6 \Omega) \text{ are series with } 2 \Omega.$ $i = i_1 + i_2 = 2 + 3 = 5 \, \text{A}$ $\text{The voltage across the } 2 \Omega \text{ resistor is,}$ $V_3 = i R_3 = 5 \times 2 = 10 \, \text{V}$

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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
}