Question

In the figure shown, what is the current (in Ampere) drawn from the battery? You are given
R₁ = 15Ω, R₂ = 10Ω, R₃ = 20Ω, R₄ = 5Ω, R₅ = 25Ω, R₆ = 30Ω, E = 15V (A) 13/24 (B) 7/18
(C) 9/32 (D) 20/3

Answer 1

$\huge\bold{(D) 20/3}$

Answer 2

Answer:

Given:

R₁ = 15Ω, R₂ = 10Ω, R₃ = 20Ω, R₄ = 5Ω, R₅ = 25Ω, R₆ = 30Ω, E = 15V

Rs₁=R₃+R₄+R₅=50Ω

1/Rp=1/Rs₁+1/R₂

Rp=25/3Ω

Rs₂=Rp+R₁+R₆

=25/3+15+30

Re=160/3Ω

so, I=V/R=15/(160/3)=9/32A

.

$i = \frac{15}{45 + \frac{25}{3} } = \frac{3}{9 + \frac{5}{3} }$

$(c) \: i = \frac{9}{32}$