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Current Question (ID: 16629)

Question:

$\text{Three resistors are connected to a } 20 \text{ V battery with a constant voltage supply.}$ $\text{One of the resistors is a variable resistor. The resistance of the variable}$ $\text{resistor is gradually increased from } 0 \, \Omega \text{ to } 5 \, \Omega.$ $\text{Which graph correctly represents how the current drawn from the battery}$ $\text{varies with the resistance } (R) \text{ of the variable resistor?}$

Options:

1. $\text{Graph 1}$
2. $\text{Graph 2}$
3. $\text{Graph 3}$
4. $\text{Graph 4}$

Solution:

$\text{Hint: } V = IR$ $\text{Step 1: Find the current through the circuit when } R = 0.$ $\text{The equivalent resistance of the circuit is given by:}$ $R_{eq} = \frac{10 \times 5}{10 + 5} = \frac{10}{3} \, \Omega$ $\text{The maximum current through the circuit when } R = 0 \text{ is given by:}$ $I = \frac{20 \times 3}{10} = 6 \, \text{A}$ $\text{Step 2: Find the current through the circuit when } R = 5 \, \Omega.$ $\text{The equivalent resistance of the circuit is given by:}$ $R'_{eq} = \frac{10 \times 10}{10 + 10} = 5 \, \Omega$ $\text{The current through the circuit when } R = 5 \, \Omega \text{ is given by:}$ $I = \frac{20}{5} = 4 \, \text{A}$ $\text{So, the current gradually decreases with resistance } R \text{ as shown in the figure below:}$ $\text{Graph 4}$ $\text{Hence, option (4) is the correct answer.}$

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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
}