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Current Question (ID: 16676)

Question:

$\text{In the given circuit diagram, 3 identical bulbs are connected. If bulb } B_3 \text{ get fused suddenly, how will the brightness of bulbs } B_1 \text{ and } B_2 \text{ change?}$

Options:

1. $\text{Brightness of bulb } B_1 \text{ will increase whereas brightness of bulb } B_2 \text{ will decrease}$
2. $\text{Brightness of bulb } B_2 \text{ will increase whereas brightness of bulb } B_1 \text{ will decrease}$
3. $\text{Brightness of both bulbs } B_1 \text{ and } B_2 \text{ will increase}$
4. $\text{Brightness of bulb } B_1 \text{ will increase whereas brightness of bulb } B_2 \text{ will remain the same}$

Solution:

$\text{Hint: If one bulb fuses off, the connection is cut and the current through that branch becomes zero.}$ $\text{Step 1: When bulb } B_3 \text{ is not fused.}$ $R_{\text{eq}} = \frac{3R}{2}$ $I = \frac{2V}{3R}$ $\text{The power dissipated in the bulb } B_1 \text{ and } B_2.$ $P_{B_1} = \left(\frac{2V}{3R}\right)^2 R = \frac{4V^2}{9R}$ $P_{B_2} = \left(\frac{V}{3R}\right)^2 R = \frac{V^2}{9R}.$ $\text{Step 2: When bulb } B_3 \text{ is fused.}$ $I_1 = \frac{V}{2R}.$ $\text{The power dissipated in } B_1 \text{ and } B_2.$ $P_{B_1} = P_{B_2} = I_1^2 R = \left(\frac{V}{2R}\right)^2 R = \frac{V^2}{4R}.$ $\text{As power dissipation in } B_1 \text{ is reduces whereas in } B_2 \text{ increases.}$ $\text{Hence, the Brightness of bulb } B_2 \text{ will increase whereas the brightness of bulb } B_1 \text{ will decrease.}$

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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
}