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

Question

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

TestHub · Accepted Answer

The equivalent resistance R e q of this given network is, R e q = 15 + 25 3 + 30 = 45 + 25 + 90 3 = 160 3 By applying ohm's law, the Current i through the battery is i = 15 160 3 = 15 × 3 160 ∴ i = 9 32 A

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