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Current Question (ID: 17367)

Question:
$\text{An inductor of } 20 \text{ mH, a capacitor of } 100 \ \mu\text{F, and a resistor of } 50 \, \Omega \text{ are connected in series across a source of emf, } V = 10 \sin(314t). \text{ What is the power loss in this circuit?}$
Options:
  • 1. $0.79 \text{ W}$
  • 2. $0.43 \text{ W}$
  • 3. $2.74 \text{ W}$
  • 4. $1.13 \text{ W}$
Solution:
$\text{As given } V_0 = 10\text{V, } \omega = 314\text{ rad/s}$ $P = V_{\text{rms}} I_{\text{rms}} \cos \phi$ $P = V_{\text{rms}} \left( \frac{V_{\text{rms}}}{Z} \right) \left( \frac{R}{Z} \right)$ $P = \frac{(V_{\text{rms}})^2 R}{Z^2}$ $\text{And } X_L = \omega L = 314 \times 20 \times 10^{-3} = 6.280$ $X_C = \frac{1}{\omega C} = \frac{1}{314 \times 100 \times 10^{-6}} = 31.84$ $\text{The impedance of the circuit is } Z = \sqrt{(X_C - X_L)^2 + R^2}$ $Z = \sqrt{(31.84 - 6.28)^2 + 50^2}$ $Z = 56 \, \Omega$ $P = \left( \frac{10}{\sqrt{2}} \right)^2 \times \frac{50}{56^2}$ $P = 0.79 \text{ W}$

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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
}