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Current Question (ID: 16727)

Question:

$\text{In the circuit shown below, if a conducting wire is connected between points } A \text{ and } B, \text{ the current in this wire will: (All resistances are given in ohms)}$

Options:

1. $\text{flow from } A \text{ to } B$
2. $\text{flow in the direction which will be decided by the value of } V$
3. $\text{be zero}$
4. $\text{flow from } B \text{ to } A$

Solution:

$\text{Hint: } V = IR$ $\text{Step: Find the current in the wire.}$ $\text{Given,}$ $\text{An electric circuit, A and B are connected with a conducting wire.}$ $\text{Current will flow from higher to lower potential.}$ $\text{Resistances } 4\Omega \text{ are connected in series, so their effective resistance is}$ $R' = 4 + 4 = 8 \Omega$ $\text{Similarly, } 1\Omega \text{ and } 3\Omega \text{ are in series}$ $\text{So, } R'' = 1 + 3 = 4 \Omega$ $\text{Now } R' \text{ and } R'' \text{ will be in parallel, hence effective resistance,}$ $R = \frac{R' \times R''}{R' + R''} = \frac{8 \times 4}{8 + 4} = \frac{32}{12} = \frac{8}{3} \Omega$ $\text{Current through the circuit, from Ohm's law}$ $i = \frac{V}{R} = \frac{3V}{8}$ $\text{Let currents } i_1 \text{ and } i_2 \text{ flow in the branches as shown.}$ $\therefore 8i_1 = 4i_2$ $\Rightarrow i_2 = 2i_1 \ldots (i)$ $\text{Also, } i = i_1 + i_2$ $\Rightarrow \frac{3V}{8} = i_1 + 2i_1$ $\Rightarrow i_1 = \frac{V}{8} \text{ A}$ $\text{And, } i_2 = \frac{V}{4} \text{ A}$ $\text{Potential drop at } A, V_A = 4 \times i_1 = \frac{4V}{8} = \frac{V}{2}$ $\text{Potential drop at } B, V_B = 1 \times i_2 = 1 \times \frac{V}{4} = \frac{V}{4}$ $\text{Since the drop of potential is greater in } 4\Omega \text{ resistance so. It will be at a lower}$ $\text{potential than } B, \text{ hence, on connecting the wire between points } A \text{ and } B, \text{ the}$ $\text{current will flow from } B \text{ to } A.$ $\text{Hence, option (4) is the correct answer.}$

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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
}