Figure shows a Wheatstone bridge circuit. Which of the following correctly shows the currents $I_{1}, I_{2}$ and $I_{3}$ in the decreasing order of magnitude?

I_{1}, I_{2}, I_{3}

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I_{2}, I_{3}, I_{1}

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I_{2}, I_{1}, I_{3}

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I_{3}, I_{2}, I_{1}

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Solution

The correct option is D $I_{2}, I_{1}, I_{3}$
Here $R_{A B} = 100 Ω, R_{B C} = 100 Ω, R_{A D} = 50 Ω, R_{D C} = 50 Ω$

$\frac{R_{A B}}{R_{B C}} = \frac{100}{100} = 1$ and $\frac{R_{A D}}{R_{D C}} = \frac{50}{50} = 1$

So $\frac{R_{A B}}{R_{B C}} = \frac{R_{A D}}{R_{D C}} \to$ bridge is balanced and no current will pass through $25 Ω$ . i.e $I_{3} = 0$ .

Here group (50 , 50) and (100 , 100) are in parallel so potential of each gruop is equal i.e $V$ (say)

thus, $I_{1} = \frac{V}{100 + 100} = V / 200$ and $I_{2} = \frac{V}{50 + 50} = V / 100$
$∴ I_{2} > I_{1} > I_{3}$

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I_{1}, I_{2} & I_{3}

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I_{1}, I_{2} & I_{3}

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