An electric lamp, whose resistance is `20 Omega` and a conductor of `4 Omega` resistance are connected to a `6 V` battery as shown in (Fig. 3.18) Calculate.
(a) the total resistance of the circuit,
(b) the current through the circuit, and
( c) the potential difference across the electric lamp and the conductor.
.

Explanation

The resistance of electric lamp, `R_(1)=20 Omega`,
The resistance of the conductor connected in series, `R_(2)=4Omega`.
Then the total resistance in the circuit
`R = R_(1)+R_(2)`
`R_(S)=20 Omega+4 Omega=24 Omega`.
The total potential difference across the two terminals of the battery
V = 6 V.
Now by Ohm's law, the current through the circuit is given by
`I = C//R_(s)`
`=6 V//24 Omega`
`=0.25 A`.
Applying Ohm's law to the electric lamp and conductor separately. we get potential difference across the electric lamp.
`V_(1)=20Omegaxx0.25 A`
=5V,
and,
`"That across the conductor,"V_(2)=4Omegaxx0.25 A`
1V.
Suppose that we like to replace the series combination of electric lamp and conductor by a single and equivalent resistor. Its resistance much be such that a potential difference of 6V across the battery terminals will cause a current of 0.25 A in the ciruit. The resistance R of this equivalent resistor would be
`R=VB//I`
`=6V//0.25A`
`=24 Omega`.
This is the total resistance of the series circuit, it is equal to the sum of the resistances.