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Current Question (ID: 16630)

Question:

$\text{A student has three } 6.0 \, \Omega \text{ resistors that can be connected together in any configuration.}$ $\text{What are the maximum and minimum resistances that can be obtained by using one or more of these three resistors?}$ $\text{(Assume the connections between the resistors have negligible resistance, the temperature of the resistors is constant, and the resistors are used in a d.c. circuit and none of the resistors are short circuited.)}$

Options:

1. $\text{maximum resistance: } 12 \, \Omega; \text{ minimum resistance: } 0.50 \, \Omega$
2. $\text{maximum resistance: } 6.0 \, \Omega; \text{ minimum resistance: } 0.50 \, \Omega$
3. $\text{maximum resistance: } 18 \, \Omega; \text{ minimum resistance: } 6.0 \, \Omega$
4. $\text{maximum resistance: } 18 \, \Omega; \text{ minimum resistance: } 2.0 \, \Omega$

Solution:

$\text{Hint: Join the resistors in series and parallel combinations.}$ $\text{Step 1: Find the maximum resistance of the resistors.}$ $\text{In series combination, the effective resistance is given by:}$ $R = R_1 + R_2 + R_3$ $\Rightarrow R_{\text{series}} = 6 + 6 + 6 = 18 \, \Omega$ $\text{Step 2: Find the minimum resistance of the resistors.}$ $\text{In parallel combination, the effective resistance is given by:}$ $\frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$ $\Rightarrow \frac{1}{R_{\text{parallel}}} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6}$ $\Rightarrow R_{\text{parallel}} = 2 \, \Omega$ $\text{Therefore, the maximum resistance of the resistor is } 18 \, \Omega \text{ and the minimum resistance of the resistor is } 2.0 \, \Omega.$ $\text{Hence, option (4) is the correct answer.}$

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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
}