BITSAT 2018 Slot 2 Solved Paper

The given circuit is equivalent to the balanced Wheatstone's network.
Let, resistance in armAB is P,P=15+6=21Ω
Resistance in armBC is Q,Q=(‌

8X

8+X

)+(3)Ω
Resistance in armDA is R,R=(‌

(6)(6)

6+6

)+(15)=3+15=18Ω
Resistance in armCD is S,S=(‌

(4)(4)

4+4

)+(4)=2+4=6Ω
Using balancing condition for Wheatstone's network.
‌

P

Q

=‌

R

S

‌

21

(‌ 8X 8+X )+(3)

=‌

18

6

‌

21

(‌ 8X 8+X )+(3)

=3
(‌

8X

8+X

)+(3)=‌

21

3

(‌

8X

8+X

)+(3)=7
‌

8X

8+X

=4
8X=32+4X
4X=32
X=8Ω