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Current Question (ID: 16662)

Question:

$\text{Power consumed in the given circuit is } P_1. \text{ On interchanging the position of } 3 \, \Omega \text{ and } 12 \, \Omega \text{ resistances, the new power consumption is } P_2. \text{ The ratio of } \frac{P_2}{P_1} \text{ is:}$

Options:

1. $2$
2. $\frac{1}{2}$
3. $\frac{3}{5}$
4. $\frac{2}{5}$

Solution:

$\text{Hint: Recall the series and parallel configuration.}$ $\text{Step 1: Find the equivalent resistance of the circuit.}$ $R_1 = 12 + \frac{6 \times 3}{6 + 3} = 14 \, \Omega$ $\text{The equivalent resistance of the circuit is given by:}$ $R_1 = 3 + \frac{6 \times 12}{6 + 12} = 7 \, \Omega$ $\text{Step 2: Find the ratio of power.}$ $\text{The ratio of power is given by:}$ $P = \frac{V^2}{R_{eq}}$ $\frac{P_2}{P_1} = \frac{R_1}{R_2} = \frac{14}{7}$ $\frac{P_2}{P_1} = 2$ $\text{Hence, option (1) is the correct answer.}$

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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
}