Question

please help for this ques

Answer 1

Given

P.D = 12 v

let R₁ = 10Ω  R₂ = 3Ω AND R₃ = 6Ω

R₂ AND R₃ ARE IN PARALLEL

1/R' = 1/R₂  + 1/R₃

1/R' = 1/3 + 1/6

1/R' = (6+3)/18

1/R' = 9/18

1/R' = 1/2

R'=2 Ω

R1 AND PARALLEL COMBINATION OF R2 AND R3 AR IN SERIES

R = 10 + 2

   = 12 Ω

USING OHMS LAW

V = IR

12 = I X 12

I = 12/12

 = 1 A

CURRENT THROUGH RESISTOR 3Ω

= R = V/I

  3 = 12/I

  I = 12/3

  I = 4 A