


The equivalent resistanceReq of this given network is,
Req=15+325+30=345+25+90=3160
By applying ohm's law, the Current i through the battery is
i=316015=16015×3
∴i=329A
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

Held on 8 Apr 2019 · Verified 6 Jul 2026.
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