Consider a Wheatstone bridge with resistance and capacitance connected as shown Find the condition on the resistance and the capacitance such that the bridge remains balanced at all times.

Explanation

Suppose that the bridge is balanced i.e. `V_(AB) = V_(AD)` and `V_(BC) = V_(DC)`Let the current and the charges on the capacitors in the circuit as shown then
`I_(1)R_(1)=I_(3)R_(3)ldot`s(i)
`(q_(2)/(C_(2))=(q_(4)/(C_(4))`ldots(ii)
Consider the charging of the part of the circuit shown alongside. Let qz be the charge on `C_(2)` and `i_(1)` be the current in the circuit. Then
`I_(1)R_(1)+q_(2)/(C_(2))=epsilonldots(iii)`
`(dq_(2))/(dt)+(q_(2))/(R_(2)C-(2))=i_(1)ldots(iv)`
Substituting (iv) in (ill) and simplifying
`(dq_(2))/(dt)+(q_(2))/(R_(eq)C_(2))=(epsilon)/(R_(1))`where`(1)/(R_(eq))=(1)/(R_(1))+(1)/(R_(2))`
therefore`q_(2)=(epsilonR_(4)C_(4))/(R_(1))[1-e^((1)/(R_(EqC_(4)))]]=(epsilonR_(2)C_(2))/R_(1)+R_(2)[1-e^(-1//R_(eqC_(2))]]`
Similarly, for the other circuit, we have`q_(4)=(epsilonR_(4)C_(4))/(R_(3)+R_(4))[1-e^(1)/(R_(eq)C_(4))]`,where`(1)/(R_(eq))=(1)/(R_(3))+(1)/(R_(4))`
NOW,`(q_(2))/(C_(2))=(q_(4)/C-(4))`,which leads to
`(R_(2))/(R_(1)+R_(2))=(R_(4))/(R_(3)+R_(4))`and`(1)/(R_(eq)C_(2))=(1)/(R_(eq)C_(2))Or(R_(1))/(R_(2))=(R_(3))/(R_(4))`and`R_(1)C_(2)=R_(3)C_(4)`