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


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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