In the figure shown, what is the current (in Ampere) drawn from the battery ? You are given: R1 = 15 Ω, R2 = 10 Ω , R3 = 20 Ω, R4 = 5 Ω , R5 = 25 Ω, R6 = 30 Ω, E = 15 V

Question

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

Tardigrade Admin · Accepted Answer

Req = 15+ (25/3) + 30 = (45+25+90/3) = (160/3) I = (E/Req) = (15×3/160) = (9/32) amp

Q. In the figure shown, what is the current (in Ampere) drawn from the battery ? You are given:
$R_{1} = 15Ω, R_{2} = 10Ω, R_{3} = 20Ω, R_{4} = 5Ω, R_{5} = 25Ω, R_{6} = 30Ω, E = 15 V$

7/18

13/24

9/32

20/3

$R_{e q} = 15 + \frac{25}{3} + 30 = \frac{45 + 25 + 90}{3} = \frac{160}{3}$
$I = \frac{E}{R _{e q}} = \frac{15 \times 3}{160} = \frac{9}{32} am p$

Q. 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