Question

1. A copper wire of resistivity p is stretched to reduce its diameter to half of its previous value. What will be its new resistivity?
2. Two bulbs of 60 W -220 V and 100 W-220 V are connected in parallel to 220 V mains. Which one of the two will glow brighter?

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Answer 1

A wire whose resistance is R is cut into 3 pieces. What is the equivalent resistance in a parallel combination?

If the pieces are of equal length, then you can think of it two ways.

First consider if instead of cutting the wire, you just folded it over on itself. Then the new arrangement would have 1/3 the length and 3 times the cross-sectional area. Resistance is proportional to the length and inversely proportional to the cross-sectional area. Therefore the resistance would be 1/3 of 1/3 of R, which is R/9.

Or do it with the parallel resistance formula. Each piece has an equal amount of the resistance, so R/3.

The formula for resistances in parallel is 1/total = 1/a + 1/b+1/c.

1/(R/3) = 3/R

1/total = 3/R + 3/R + 3/R = 9/R

So total = R/9

Answer 2

Hey mate!!

_________★_★

1) RESISTIVITY is independent of change in dimensions,

so it will remain same.

2)

★ For series combination

BRIGHTNESS IS INVERSELY PROPORTIONAL TO RATING

SO, bulb of less wattage (60W) will glow more

★ for PARALLEL combination

BRIGHTNESS IS DIRECTLY PROPORTIONAL TO RATING

SO, Bulb of high wattage(100W ) will glow more.

hope it helps!!

#Jaihind