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Current Question (ID: 17380)

Question:
$\text{A transformer has an efficiency of } 90\% \text{ when working on a } 200 \text{ V and } 3 \text{ kW power supply. If the current in the secondary coil is } 6 \text{ A, the voltage across the secondary coil and the current in the primary coil, respectively, are:}$
Options:
  • 1. $300 \text{ V, } 15 \text{ A}$
  • 2. $450 \text{ V, } 15 \text{ A}$
  • 3. $450 \text{ V, } 13.5 \text{ A}$
  • 4. $600 \text{ V, } 15 \text{ A}$
Solution:
$\text{Current in primary coil} = \frac{P}{V} = \frac{3000}{200} = 15 \text{ A}$ $\text{Efficiency is } 90\%$ $\text{So, output power is } 90\% \text{ of input}$ $P_o = \left(\frac{90}{100}\right) \times 3000$ $\text{If output voltage and current - } V_o I_o$ $V_o I_o = \left(\frac{90}{100}\right) \times 3000$ $V_o \times 6 = \left(\frac{90}{100}\right) \times 3000$ $V_o = 450 \text{ V}$

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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
}