Answer :

Here a battery of 9 V is connected in series with resistors of R1=0.2ohm, R2=0.4ohm, R3=0.3ohm, R4=0.5ohm, R5=12ohm,

So the resultant resistance = R1 + R2 + R3 + R4 + R5


R = 0.2+0.4+0.3+0.5+12=13.4ohm


As we know that


V= IR


Thus the current flow through 12ohm resistance will be equal to the current flowing across the whole circuit as in case of Series Connection same current flows through each resistance


∴ I = V/R


I = 9/13.4


I = 0.67amp.

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