(a) Three resistors of resistance `2Omega,3Omega and 4Omega` are combined in series. What is the total resistance of the combination ?
(b) It this combination is connected to a battery of emf 10 V and negligible internal resistance, obtain the potential drop across each resistor.

Explanation

(a) `R_(eq)=R_(1)+R_(2)+R_(3)=2+3+4=9Omega`
(b) `V=IR_(eq),I=(V)/(R_(eq))=(10)/(9)=1.11A`
finding potential drop across, `R_(1),V_(1)=IR_(1)=2.22V`
finding potential drop across `R_(2),V_(2)=IR_(2)=3.33V`
finding potential drop across `R_(3),V_(3)=IR_(3)=4.44V` Detailed answer:
`R_(1)=2Omega,V=10` volts
`R_(3)=3Omega`
`R_(4)=4Omega`
`R_(S)=R_(1)+R_(2)+R_(3)=2+3+4=9Omega`
but `I=(V)/(R_(S))=(10)/(9)=1.1`A
`therefore V_(1)=IR_(1)=1.1xx2=2.2` volts
`V_(2)=IR_(2)=1.1xx3=3.3` volts
`V_(3)=IR_(3)=1.1xx4=4.4` volts