a battery of 9v is connected in series with resistors of 0.2, 0.3, 0.4,0.5 and 12 ohm respectivly. how much current would flow through the 12 ohm resistor?
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Answer:
potential difference = 9v
Total resistance in series = 0•2+0•3+0•4+0•5 +12
= 1•4+12 = 13•4 ohm
Total current = V/ R = 9 / 13.4
=0.671641791044776 ≈ 0.67 A
As current in series is same
so current through 12 ohm resistor will also 0.67 Ampere
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