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Current Question (ID: 20285)

Question:

$\text{A circuit to verify Ohm's law uses ammeter and voltmeter in series or parallel connected correctly to the resistor. In the circuit:}$

Options:

1. $\text{The ammeter is always connected in series and the voltmeter is in parallel.}$
2. $\text{Both the ammeter and voltmeter must be connected in series.}$
3. $\text{Both the ammeter and voltmeter must be connected in parallel.}$
4. $\text{The ammeter is always connected in parallel and the voltmeter is in series.}$

Solution:

$\text{Hint: An ammeter is used to measure the current flowing through the circuit and hence, it is connected in series.}$ $\text{Explanation: Ammeter: In a series connection, the same current flows through all the components. It aims at measuring the current flowing through the circuit so, the ammeter is connected in series.}$ $\text{Voltmeter: A voltmeter measures voltage change between two points in a circuit, so we have to place the voltmeter in parallel with the circuit components.}$ $\text{This configuration ensures that the ammeter measures the current flowing through the circuit (in series), while the voltmeter measures the voltage across the resistor (in parallel), allowing you to calculate the resistance using Ohm's law } \left( R = \frac{V}{I} \right).$

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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
}