Identify correct relation for the circuit mentioned below, assuming the current $I_{3}$ flows through a conductor with zero resistance:

I_{1} = I_{2} + I_{3} + I_{4}

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I_{1} = I_{3}

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I_{1} = I_{2}

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I_{3} = I_{5}

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Solution

The correct options are
A $I_{1} = I_{2} + I_{3} + I_{4}$
B $I_{1} = I_{3}$
D $I_{3} = I_{5}$
If an ideal conductor (i.e. a conductor without any resistance) is connected across a circuit, all the current will flow through the conductor.
$∴ I_{2} = I_{4} = 0$
Also, $I_{1} = I_{2} + I_{3} + I_{4}$
$\Rightarrow I_{1} = I_{3}$
We know that in series connection current remains the same.
$∴ I_{1} = I_{2} + I_{3} + I_{4} = I_{3} = I_{5}$

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Similar questions

Now what should be the correct relation between $I_{1}$ , $I_{2}$ and $I_{3}$ .

A B C D

O

1, 2, 3

4

I_{1}, I_{2}, I_{3} & I_{4}

I_{1} = 1 \int 0 e^{- x} {cos}^{2} x d x

I_{2} = 1 \int 0 e^{- x^{2}} {cos}^{2} x d x

I_{3} = 1 \int 0 e^{- x^{3}} d x

I f I_{1} = \int_{0}^{1} 2^{x^{2}} d x, I_{2} = \int_{0}^{1} 2^{x^{3}} d x, I_{3} = \int_{1}^{2} 2^{x^{2}} d x a n d I_{4} = \int_{1}^{2} 2 x^{3} d x t h e n

I_{1} = \int_{\frac{π}{3}}^{\frac{π}{6}} \frac{sin x}{x} d x, I_{2} = \int_{\frac{π}{3}}^{\frac{π}{6}} \frac{sin (sin x)}{sin x} d x, I_{3} = \int_{\frac{π}{3}}^{\frac{π}{6}} \frac{sin (tan x)}{tan x} d x

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