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>> $Consider the circuit shown in the figure$

Question

Consider the circuit shown in the figure. Assuming $R_{2} = 2 Ω$ , $R_{3} = 4 Ω a n d R_{4} = 2 Ω$ current passing through resistance $R_{1}$ is:

158567

A

$2 A$

B

$2.5 A$

C

$3.5 A$

D

$0 A$

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Updated on : 2022-09-05

Solution

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Correct option is B)

BY KIrchhoff's voltage law,

$15 - I R_{3} - I R_{4} = 0$

      Given $R_{3} = 4 Ω$ and $R_{4} = 2 Ω$

      $I = 2.5 A$ through $R_{3}$ and $R_{4}$

     $- 10 - I R_{1} + 5 = 0$

or $I = \frac{5}{R _{1}} = 2.5 A$

     $15 - 6 I - 5 - 2 I + 10 = 0$

or $I = \frac{2 0}{8} = 2.5 A$

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