NCERT Solution for Class 10 Science Chapter 12

Electricity

NCERT Books for Session 2020-2021

CBSE Board and UP Board

Exercises Questions

Page No-221

Questions No-12

# Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?

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For one bulb: Power P = 10 W and Potential Difference V = 220 V

Using the relation for R, we have

𝑅 = 𝑉²/𝑃= (220) ²/10 = 4840 Ω

Let the total number of bulbs be x.

Given that: Current I = 5 A and Potential Difference V = 220 V

According to Ohm’s law, V = IR

⟹𝑅=𝑉/𝐼=220/5=44 Ω

Now, for x number of bulbs of resistance 176 Ω, the equivalent resistance of the resistors connected in parallel is 44 Ω.

1/44=1/4840 + 1/4840 + 1/4840 + ..𝑡𝑜 𝑥 𝑡𝑖𝑚𝑒𝑠 ⟹1/44=𝑥/4840 ⟹𝑥=484044=110

Therefore, 110 bulbs of 4840 Ω are required to draw the given amount of current.

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https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/

Here, current, I = 5 A, voltage, V = 220 V

∴ Maxium power, P = I x V = 5 x 220 = 1100W

Required no. of lamps =Max.Power/Powerof1lamp=1100/10=110

∴ 110 lamps can be connected in parallel.