AP EAPCET 21 May 2025 Shift 1 Paper

‌R1=100Ωl1=l
‌‌‌‌‌R2‌=?,l2=l1+20%‌ of ‌l1
‌=l+0.2l=1.2l
∴‌‌R′‌=n2R=(1.2)2R
‌=1.44R=1.44×100=144Ω
The wire of resistance 144Ω is bent into a rectangle with sides in the ratio 3:2. Let the side be 3x and 2x.
Total length, l=2(3x+2x)=10x
∴‌‌R‌length ‌‌=‌

3x

10x

×144=43.2Ω
R‌width ‌‌=‌

2x

10x

×144=28.8Ω
The effective resistance across the diagonal is the equivalent resistance of two parallel branches. On branch consists of the other two sides of the same resistance. Each branch has a resistance of
‌R‌length ‌+R‌width ‌=43.2+28.8=72Ω
‌∴‌‌‌

1

R‌eq ‌

=‌

1

72

+‌

1

72

=‌

1

36

‌⇒‌‌R‌eq ‌=36Ω