In a Wheatstone's bridge, three resistances P, Q and R are connected in the three arms and fourth arm is formed by two resistances S1, and S2, connected in parallel. The condition for the bridge to be balanced will be
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a
PQ=2RS1+S2
b
PQ=RS1+S2S1S2
c
PQ=RS1+S22S1S2
d
PQ=RS1+S2
answer is B.
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Detailed Solution
See fig.For balanced Wheatstone's bridgePQ=RS…………(1)Here S=S1∥S2Substituting the value of S from eq. (2) in eq. [1), we getPQ=RS1S2/S1+S2=RS1+S2S1S2