Page 189 - NCERT Science Class 10 English Medium

On applying Ohm’s law to the three resistors separately, we further
have
V = I R [11.13(a)]
1 1
V = I R [11.13(b)]
2 2
and V = I R [11.13(c)]
3 3
From Eq. (11.11),
I R = I R + I R + I R
1 2 3
or
R = R +R + R (11.14)
s 1 2 3
We can conclude that when several resistors are joined in series, the
resistance of the combination R equals the sum of their individual
s
resistances, R , R , R , and is thus greater than any individual resistance.
1 2 3
Example 11.7
An electric lamp, whose resistance is 20 Ω, and a conductor of 4 Ω
resistance are connected to a 6 V battery (Fig. 11.9). 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.

Figure 11.9
Figure 11.9 An electric lamp connected in series with
Figure 11.9
Figure 11.9
Figure 11.9
a resistor of 4 Ω to a 6 V battery
Solution
The resistance of electric lamp, R = 20 Ω,
1
The resistance of the conductor connected in series, R = 4 Ω.
2
Then the total resistance in the circuit
R = R + R
1 2
R = 20 Ω + 4 Ω = 24 Ω.
s
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 = V/R
s
= 6 V/24 Ω
= 0.25 A.

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