An electric lamp, whose resistance is 20 Omega, and a conductor of 4 Omegaresistance are connected to a 6 V battery (see figure).
Calculate (a) the total resistance of the circuit, (b) the current through the circuit, and (c ) the potential difference across the electric lamp and conductor.

Solution :( a) The resistance of electric lamp, `R_1 = 20 Omega`
The resistance of the conductor connectedin series,`R_2 = 4Omega`
Then the TOTAL resistance in the circuit
`R = R_1 + R_2`
`R_s = 20 Omega + 4Omega =24 Omega`
( b )The total POTENTIAL difference across the two terminals of the battery
`V = 6V`
Now by the Ohm.s law, the current through the circuit is given by
`I = V/R_a`
` = 6 V/24 Omega`
`=0.25 A`
( C )APPLYING the Ohm.s law to the electric lamp and conductor separately, we get potential difference across the electric lamp,
`V_1 = 20 Omega xx 0.25 A`
`= 5V`
and, that across the conductor,
`V_2 = 4Omega xx 0.25 A = 1V`
Note: Suppose, we like to replace the series combination of electric lamp and conductor by a single and equivalent resistor. Its resistance must be such that a potential difference of 6 V acrosS the battery terminals will cause a current of 0.25 A in the circuit. The resistance R of this equivalent resistor would be
`R =V/I`
` = 6V //0.25 A`
`=24 Omega`
This is the total resistance of the series circuit, it is equal to the sum of the two resistances.