Electricity

Activity . In Activity . , insert a voltmeter across the ends X and Y of the series combination of three resistors, as shown in Fig. .

. Plug the key in the circuit and note the voltmeter reading. It gives the potential difference across the series combination of resistors. Let it be V .

Now measure the potential difference across the two terminals of the battery. Compare the two values. Take out the plug key and disconnect the voltmeter. Now insert the voltmeter across the ends X and P of the first resistor, as shown in Fig.

. . Figure . Figure .

Figure . Figure . Figure . Plug the key and measure the potential difference across the first resistor.

Let it be V . Similarly, measure the potential difference across the other two resistors, separately. Let these values be V and V , respectively. Deduce a relationship between V , V , V and V .

You will observe that the potential difference V is equal to the sum of potential differences V , V , and V . That is the total potential difference across a combination of resistors in series is equal to the sum of potential difference across the individual resistors. That is, V = V + V + V ( . ) In the electric circuit shown in Fig.

. , let I be the current through the circuit. The current through each resistor is also I . It is possible to replace the three resistors joined in series by an equivalent single resistor of resistance R , such that the potential difference V across it, and the current I through the circuit remains the same.

Applying the Ohm’s law to the entire circuit, we have V = I R ( . ) On applying Ohm’s law to the three resistors separately, we further have V = I R [ . (a)] V = I R [ . (b)] and V = I R [ .

(c)] From Eq. ( . ), I R = I R + I 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