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

Question

In the figure shown, what is the current (in Ampere) drawn from the battery? You are given:

$R_{1} = 15Ω, R_{2} = 10Ω, R_{3} = 20Ω, R_{4} = 5Ω, R_{5} = 25Ω, R_{6} = 30Ω, E = 15 V$

TestHub · Accepted Answer

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

Physics - Current Electricity Question with Solution | TestHub

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