CGPET 2016 Solved Paper

Equivalent resistance of loop I,
=2+4+3=8Ω
Req=

R1×R2

R1+R2

=

8×8

8+8

=

8×8

16

=4Ω
Equivalent resistance of loop II,
=2+4+2=8Ω
R′eq=

R1×R2

R1+R2

=

8×8

8+8

=4Ω
Equivalent resistance loop III,
=3+2+4=9Ω
Maximum current, i=

6

9

=

2

3

A
=0.66A
This current is divided into two equal parts.
Current in first 2 Ω resistor =

1

3

A
The current in 4Ω resistor =

1

3×2

A
=

1

6

A=0.167A