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Current Question (ID: 20252)

Question:

$\text{The current through a } 5 \, \Omega \text{ resistance remains the same, irrespective of its connection across a series or parallel combination of two identical cells. The internal resistance of the cell is:}$

Options:

1. $5 \, \Omega$
2. $10 \, \Omega$
3. $15 \, \Omega$
4. $20 \, \Omega$

Solution:

$\text{Hint: Recall the series and parallel configuration of cells.}$ $\text{Step 1: Find the internal resistance and EMF of the cell when they are connected in parallel.}$ $\text{When connected in parallel (case: A)}$ $\text{Effective resistance is given by: } \frac{1}{r_{eq}} = \frac{1}{r} + \frac{1}{r} \Rightarrow r_{eq} = \frac{r}{2}$ $\text{Effective EMF is given by: } \frac{\varepsilon_{eq}}{r/2} = \frac{\varepsilon}{r} + \frac{\varepsilon}{r} \Rightarrow \varepsilon_{eq} = \varepsilon$ $\Rightarrow \text{current, } i_A = \frac{\varepsilon}{R + \frac{r}{2}}$ $\text{Step 2: Find the internal resistance and EMF of the cell when they are connected in series.}$ $\text{When connected in series (case: B)}$ $\text{Effective EMF is given by: } \varepsilon_{eq} = \varepsilon + \varepsilon = 2\varepsilon$ $\text{Effective resistance is given by: } r_{eq} = r + r = 2r$ $\Rightarrow \text{current, } i_B = \frac{2\varepsilon}{R + 2r}$ $\text{As, } i_A = i_B$ $\Rightarrow \frac{\varepsilon}{R + \frac{r}{2}} = \frac{2\varepsilon}{R + 2r}$ $\Rightarrow R + 2r = 2R + r$ $\Rightarrow R = r = 5 \, \Omega$ $\text{Hence, option (1) is the correct answer.}$

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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
}