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Given Battery=9Vresistors of 0.2 ohm, 0.3ohm, 0.4ohm, 0.5ohm and 12ohm are connected in series to battery.R1=0.2ohmR2=0.3ohmR3=0.4ohmR4=0.5ohmR5=12ohmAs all are connected in series effective resistance would be :Reff=R1+R2+R3+R4+R5=0.2+0.3+0.4+0.5+12=13.4OHMS
By ohms law : V=IRI=V/R=9/13.4=0.67 A
As resistors are connected in series. I=I1=I2=I3=I4=I5Hence current across 12ohms resistor is 0.67A
By ohms law : V=IRI=V/R=9/13.4=0.67 A
As resistors are connected in series. I=I1=I2=I3=I4=I5Hence current across 12ohms resistor is 0.67A
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