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Question

A battery of 9 V is connected in series with resistors of 0.2 Ω,0.3 Ω,0.4 Ω,0.5Ω and 12 Ω respectively. How much current will flow through the 12 Ω resistor?

Solution
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For resistors in series,
Req=R1+R2+R3+R4+R5
=0.2+0.3+0.4+0.5+12
=13.4 Ω

By ohm's Law:
V=IReq
9=13.4I
I=0.67 A

When resistors are connected in series, current is same in all the resistors. Hence, current in 12 Ω resistor =0.67 A.

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