Test: Combination of Resistors - JEE MCQ

Three 2Ω resistors are in series. Their total resistance =6Ω. Now it is in parallel with 2Ω resistor, so total resistance,
1/R​=1/2+1/6​=3+1/6​=4/6=2/3
R=3/2​
∴I=RV​=3/(3/2)​=3×2​/3=2A

In the given circuit, the arrangement of resistances is a form of Wheatstone's bridge. Hence no current will flow through the 7Ω resistor. The resistances between A and B will be a parallel combination of each other.
Hence,
1/R_eq​​=(1/2+3)​+(1/4+6)​
1/R_eq​​=(1/5)+(1/10)
∴R_eq​=(5×10/15)​
∴R_eq​=(10​/3) Ω
 

Here 4 Ω, 12 Ω, 6 Ω when connected in parallel results in 2Ω. to reduce it to half we have to join 1\R original = 6\12 for reducing it to half we have to join 6 , 12 Ω  resistors in parallel (6\12) + (1\ 12 × 6) = 12\12 = 1 ohm . Half of its original value therefore option a is correct.

We know that,
I=V/R​
∴​I₁/I₂ ​​= ​R₂/ R₁
​​R₁​=90Ω​
Current flowing through 90Ω resistance is reduced by 90%
∴ Current ratio =10%:90%
=1:9
∴1/9​=R₂​R₁
​​⇒1/9​= R₂​/90
​⇒R₂​=90/9​=10
∴R₂​=shunt=10Ω