Four resistors `R_(1),R_(2),R_(3)` and `R_(4)` are connected between two terminals A and B iin a network as shown in the diagram. A key K can connect the two points (see diagram) C and D. A constant potential difference `(V_(A)-V_(B))=V` is maintained between thhe points A and B
Q. In on connecting the key K, no current flows through the segment CD, whatever potential difference V exists between A and B which of the following cases will satisfy this condition?

Options

A. `R_(1)=1Omega,R_(2)=2Omega,R_(3)=3Omega,R_(4)=4Omega`
B. `R_(1)=2Omega,R_(2)=1Omega,R_(3)=4Omega,R_(4)=2Omega`
C. `R_(1)=1Omega,R_(2)=2Omega,R_(3)=2Omega,R_(4)=1Omega`
D. `R_(1)=R_(2)=2Omega,R_(3)=2R_(4)=4Omega`

Answer

B

Explanation

Let the resistance (small but zero) in the arm CD be x `Omega`. Assume currents `i_(1),i_(2),i_(3),i_(4)` and `i_(x)` as shown in the diagram. Then applications of kirchhoff's junctions and loop laws give,
`i_(1)=i_(2)+i_(x)` ..(i)
`i_(3)=i_(4)-i_(x)` ..(ii)
`V=i_(1)R_(1)+i_(1)R_(2)+i_(4)R_(4)` ..(iii)
and `i_(1)R_(1)+X i_(x)-i_(3)R_(3)=0` ..(iv)
Solving these equation for `i_(x)` we get from (1) & (3)
`V-i_(1)(R_(1)+R_(2))-i_(x)R_(2)` ..(v)
and from (2) and (3)
`V-i_(3)(R_(3)+R_(4))+i_(x)R_(4)` ...(vi)
Sustituting for `i_(1)` from (5) and `i_(3)` from (vi) and (iv)
`R_(1)[(V+i_(x)R_(2))/(R_(1)+R_(2))]+X i_(2)-R_(3)[(V-i_(x)R_(4))/(R_(3)+R_(4))]=0`
Collecting thew terms in
`i_(x),i_(x)[x+(R_(1)R_(2))/(R_(1)+R_(2))+(R_(3)R_(4))/(R_(3)+R_(4))]`
`=V[(R_(3))/(R_(3)+R_(4))-(R_(1))/(R_(1)R_(2))]`
`impliesI_(x)=(V[(R_(3))/(R_(3)+R_(4))-(R_(1))/(R_(1)+R_(2))])/(x+(R_(1)R_(2))/(R_(1)+R_(2))+(R_(3)R_(4))/(R_(3)+R_(4)))`
Proceeding to limit `xto0` (`x=` m negligibly small)
`i_(x)=(25[(R_(3))/(R_(3)+r_(4))-(R_(1))/(R_(1)+R_(2))])/([(R_(1)R_(2))/(R_(1)+R_(2))+(R_(3)R_(4))/(R_(3)+R_(4))])` substituting values,
`i_(x)=(25[(3)/(7)-(1)/(3)])/([(2)/(3)+(12)/(7)])=(25xx2)/(50)=`lamp