In a typical wheatstone network the resistance in cyclic order are A=`10Omega,B=5O,ega,C=4Omega` and `D=4Omega` for the bridge to be balanced.

Options

A. `10Omega` should be connected in parallel with A
B. `10Omega`should be connected in series with A
C. `5Omega` should be connected in series with B
D. `5Omega` should be connected in parallel with B

Answer

A

Explanation

Resistance in upper arms, `A =10 Omega, B = 5 Omega`
Resistance in lower arms, `C = 4 Omega, D = 4 Omega`
The condition required to form a Wheatstone bridge is given by `" "(A)/(B)=(C )/(D)`
which does not satisfy this case.
If a `10 Omega` resistance is connected in parallel with A [as per option (a)]
The relation for a new resistance is given by
`(1)/(A')=(1)/(10)+(1)/(10)=(2)/(10)=(4)/(4)=(1)/(5)`
`A'=5 Omega`
Now, the network satisfies the condition of Wheatstone bridge
`(P')/(5)=(R')/(5)` or `(5)/(5)=(4)/(4)`