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Q.

A number of 110 volt lamps connected in parallel are fed by a 130 V battery having an internal resistance of 2.6 ohm. If the resistance of each lamp is 200 ohm and the resistance of conducting wires is O .40 ohm, the number of lamps which the battery can supply is

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a

5

b

1

c

12

d

15

answer is C.

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Detailed Solution

Let n lamps be connected in parallel The current in the main circuit
$i = \frac{130}{[(200 / n) + 2 \cdot 6 + 0 \cdot 4]}$
Current through each lamp
$\frac{i}{n} = \frac{130}{n [(200 / n) + 3]}$
Potential drop across each lamp is [(i / n) x.R]. Hence
$\frac{130}{n [(200 / n) + 3]} \times 200 = 110$
Solving for n, we get n = 12

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