R + I R or R s = R +R + R ( . ) We can conclude that when several resistors are joined in series, the resistance of the combination R s equals the sum of their individual resistances, R , R , R , and is thus greater than any individual resistance. Example . An electric lamp, whose resistance is Ω , and a conductor of Ω resistance are connected to a V battery (Fig.
. ). 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. The resistance of electric lamp, R = Ω , The resistance of the conductor connected in series, R = Ω .
Then the total resistance in the circuit = R + R R s = Ω + Ω = Ω . The total potential difference across the two terminals of the battery V = V. Now by Ohm’s law, the current through the circuit is given by V / R s V/ Ω . A.
Figure . Figure . Figure . Figure .
Figure . An electric lamp connected in series with a resistor of Ω to a V battery Applying Ohm’s law to the electric lamp and conductor separately, we get potential difference across the electric lamp, V = Ω × . A = V; and, that across the conductor, V = Ω × . A = V.
Suppose that 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 V across the battery terminals will cause a current of . A in the circuit. The resistance R of this equivalent resistor would be = V / I = V/ .
A = Ω . This is the total resistance of the series circuit; it is