In a balanced Wheatstone bridge,the current i | vedclass

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(C) Wheatstone bridge is balanced when the ratio of resistances in the arms is equal,i.e.,$P/Q = R/S$. $[1]$ The balance condition is independent of t

Vedclass · Accepted Answer

$[1], [3]$ and $[4]$ are correct

Vedclass · Answer

(C) Wheatstone bridge is balanced when the ratio of resistances in the arms is equal,i.e.,$P/Q = R/S$. $[1]$ The balance condition is independent of the battery $EMF$. Changing the $EMF$ only changes the magnitude of the current in the branches but not the ratio,so the galvanometer current remains zero. $[2]$ If all resistances are increased by $10 \ \Omega$,the new ratio becomes $(P+10)/(Q+10)$ and $(R+10)/(S+10)$. This is generally not equal to $P/Q$ unless $P=Q=R=S$. Thus,this is not always true. $[3]$ If all resistances are multiplied by a factor $k=5$,the new ratio becomes $(5P)/(5Q) = P/Q$ and $(5R)/(5S) = R/S$. Since $P/Q = R/S$,the bridge remains balanced. $[4]$ According to the reciprocity theorem for a Wheatstone bridge,interchanging the battery and the galvanometer does not affect the balance condition. Therefore,statements $[1], [3],$ and $[4]$ are correct.

In a balanced Wheatstone bridge,the current in the galvanometer is zero. It remains zero when:

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