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 = 5Ω, R5 = 25 Ω, R6 = 30 Ω, E = 15 V

Engineering

Physics

Ohm Law and Current Basics

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 = 5Ω, R₅ = 25 Ω, R₆ = 30 Ω, E = 15 V

9/32

13/24

20/3

7/18

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$R_{eq} = 15 + \frac{25}{3} + 30 = \frac{45 + 25 + 90}{3} = \frac{160}{3}$

$I = \frac{E}{R_{eq}} = \frac{15 \times 3}{160} = \frac{9}{32} amp .$

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