(a) Three resistors `1 Omega, 2Omega` and `3 Omega` are combined in series. What is the total resistance of the combination ?
(b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.

Explanation

(a) The resistors of resistances `1 Omega,2 Omega and 3 Omega` are combined in series. Total resistance of the combination is given by the algebraic sum of individual resistances. ltbgt The resistances `=1+2+3=6 Omega`
(b) Current flowing through the circuit `= I`
Emf of the battery, `E = 12V`
Total resistance of the circuitm `R = 6 Omega`
The relation for current using Ohm's law is,
`I=(E)/(R)`
`=(12)/(6)=2A`
Potential drop acorss `1 Omega` resistor `= V_(1)`
From Ohm's law, the value of `V_(1)` can be obtained as
`V_(1)=2xx1=2V`...(i)
Potential drop across `2 Omega` resistor `= V_(2)`
Again, from Ohm's law, the value of `V_(2)` can be obtained as
`V_(2)=2xx2 = 4V`....(ii)
Potential drop acorss `3 Omega` resistor `= V_(3)`
Again, from Ohm's law, the value of `V_(3)` can be obtained as
`V_(3)=2xx3=6 V`....(iii)
Thereforem the potential drop across `1 Omega,2 Omega and 3 Omega` resistors are `2 V,4V and 6V` respectively.