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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. |
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Answer» Solution :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)` 
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