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

Explanation

(a) Total resistance in series combinatin of three resistors of resistances `R_(1), R_(2) and R_(3)` is given by: `R_(S) = R_(1) + R_(2) + R_(3)`
Here, `R_(1) =1 Omega, R_(2) = 2 Omega, R_(3) = 3 Omega`
`:. R_(S) = (1 +2 +3) Omega= 6Omega`
(b) The current flowing through the circuit when the combination is connected to a battery of emf E and internal resistance r is given by: `I = (E )/(R_(S) +r)`
Here, E= 12V
r=0
`:. I= (12)/(6) = 2A`
The potential drop across each resistor is shown in the following figure.

`:. V_(1) = IR_(1) = 2 xx 1= 2V`
`V_(2) = IR_(2) = 2 xx 2= 4V`
`V_(3) = IR_(3) = 2 xx 3 = 6V`