Question

a copper wire resistance of 30ohm is cut into three equal parts and each part is stretched to twice of its length find the resistance of each part

Answer 1

Answer:

HEY

When a resisteor is stretched to double in original length the new resistance becomes four times its original resistance becomes R∝l2

∴ New resistance =4R=4×15=60Ω

Resistance of each part =30Ω

i.e. R1=30ΩandR2=30Ω

When they are conpected in parallel the effective

resistance, RP=

R1R2

R1+R2

 

=

30×30

30+30

 

=15Ω

Current, I=

V

RP

 

=

3

15

 

=0.2A

Explanation:

pLZ MARK AS  BARINLIEST

Answer 2

Answer:

see the above attachment