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Current Question (ID: 16656)

Question:

$\text{The total power dissipated in watts in the circuit shown below is:}$

Options:

1. $16 \text{ W}$
2. $40 \text{ W}$
3. $54 \text{ W}$
4. $4 \text{ W}$

Solution:

$\text{Hint: The power depends on the equivalent resistance of the circuit.}$ $\text{Step 1: Find the equivalent resistance of the circuit.}$ $\frac{1}{R_1} = \frac{1}{6} + \frac{1}{3} = \frac{1+2}{6} = \frac{1}{2}$ $\text{Or } R_1 = 2 \ \Omega$ $\text{Again, } R_1 \text{ is in series with } 4 \ \Omega \text{ resistance, hence}$ $R = R_1 + 4 = 2 + 4 = 6 \ \Omega$ $\text{Step 2: Find the power dissipated.}$ $\text{Thus, the total power dissipated in the circuit,}$ $P = \frac{V^2}{R}$ $\text{Here, } V = 18 \text{ V, } R = 6 \ \Omega$ $\text{Thus, } P = \frac{(18)^2}{6} = 54 \text{ W}$

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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
}