Current $I$ in the network shown in figure is

1 A

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0.5 A

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2 A

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5 A

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Solution

The correct option is A $1 A$
(R= Resistance)
$R_{1} & R_{2} \Rightarrow$ series
$R_{e q} = R_{1} + R_{2} \Rightarrow R_{e q} = 10 + 6 = 16 Ω$ ....(i)
$R_{e q}$ & $R_{3} \Rightarrow$ Paralal
$\frac{1}{R_{(e q)_{2}}} = \frac{1}{R_{(e q)_{1}}} + \frac{1}{R_{3}} \Rightarrow \frac{1}{R_{(e q)_{2}}} = \frac{1}{16} + \frac{1}{16} \Rightarrow R_{e q_{2}} = 8$

$R_{e q_{2}}$ & $R_{4} =$ Series

$R_{e q_{3}} = R_{e q_{2}} + R_{4} = 8 + 2 = 10$

using ohm's law
$\Rightarrow V = I R_{(e q)_{3}} \Rightarrow 10 = I (10)$
hence $I = 1 A$

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R_{1} = R_{2} = R_{3} = 10 Ω

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