A battery of 15 Volt is connected to a parallel grouping of resistance 9 Ω, 18 Ω and 6 Ω. Find the
current drawn through the battery.
Answers
Answered by
1
Explanation:
Resistance = volt / current
current = volt / resistance
Given resistance are 9, 18, 6
current = 15/9 = 5/3 Ampere
current = 15/18 = 5/6
Current = 15/6 = 5/2
Answered by
0
Explanation:
net R=(9×18×6)/9+18+6
R=324/11
V/R=I
=15×11/324
I=55/108ampere
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