Question

A wire of resistance 'R' is stretched by 50%. What will be percentage change in it's resistance . The resistance of 100 w bulb is less than resistance of 40 w bulb . Explain the reason

Answer 1

Let orig length =l
New length(l*)=l+50%l=1.5l
Volume remains constant
A*l*=Al
A*/A=10/15
BECAUSE
R=(RHO ×l)/A
So R*/R=225/100
Therefore percentage change in resistance is
(R*-R)/R×100=(2.25R-R)/R×100=125
Also we know that P=V2/R
So for 100 w bulb R1 =V2//P =V2/100=0.01 V2
And for 40 w bulb
R2=V2/40=0.025V2
So resiatance of 40 w bulb is higher than resistance of 100 w bulb

Answer 2

Answer:

Explanation:

Let orig length =l

New length(l*)=l+50%l=1.5l

Volume remains constant

A*l*=Al

A*/A=10/15

BECAUSE

R=(RHO ×l)/A

So R*/R=225/100

Therefore percentage change in resistance is

(R*-R)/R×100=(2.25R-R)/R×100=125

Also we know that P=V2/R

So for 100 w bulb R1 =V2//P =V2/100=0.01 V2

And for 40 w bulb

R2=V2/40=0.025V2

So resiatance of 40 w bulb is higher than resistance of 100 w bulb