Page 191 - NCERT Science Class 10 English Medium

n Take out the plug from the key. Remove the ammeter and voltmeter from the circuit.
Insert the ammeter in series with the resistor R , as shown in Fig. 11.11. Note the ammeter
1
reading, I .
1

Figure 11.1111.11
Figure 11.11
Figure 11.11
Figure
Figure 11.11
n Similarly, measure the currents through R and R . Let these be I and I , respectively.
2 3 2 3
What is the relationship between I, I , I and I ?
1 2 3
It is observed that the total current I, is equal to the sum of the
separate currents through each branch of the combination.
I = I + I + I (11.15)
1 2 3
Let R be the equivalent resistance of the parallel combination of
p
resistors. By applying Ohm’s law to the parallel combination of resistors,
we have
I = V/R (11.16)
p
On applying Ohm’s law to each resistor, we have
I = V /R ; I = V /R ; and I = V /R (11.17)
1 1 2 2 3 3
From Eqs. (11.15) to (11.17), we have
V/R = V/R + V/R + V/R
p 1 2 3
or
1/R = 1/R + 1/R + 1/R (11.18)
p 1 2 3
Thus, we may conclude that the reciprocal of the equivalent resistance
of a group of resistances joined in parallel is equal to the sum of the
reciprocals of the individual resistances.
Example 11.8
In the circuit diagram given in Fig. 11.10, suppose the resistors R ,
1
R and R have the values 5 Ω, 10 Ω, 30 Ω, respectively, which have
2 3
been connected to a battery of 12 V. Calculate (a) the current through
each resistor, (b) the total current in the circuit, and (c) the total circuit
resistance.

Solution
R = 5 Ω, R = 10 Ω, and R = 30 Ω.
1 2 3
Potential difference across the battery, V = 12 V.
This is also the potential difference across each of the individual
resistor; therefore, to calculate the current in the resistors, we use
Ohm’s law.
The current I , through R = V/ R
1 1 1
I = 12 V/5 Ω = 2.4 A.
1

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