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Question

Derive the relation between the resistances of the arms of a Wheatstone bridge in its balance condition. Write its two applications.

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Solution

Look at the diagram above This is a wheatstone bridge.
You can notice four resistances P,Q,R and S connected on the bridge.
You can also notice a galvanometer connected between points B and D.
The four resistances are adjusted in such a way that the current through the galvanometer becomes zero.
This shows that B and D are at the same electric potential.
VB=VD
The potential difference across B will be VAVB
Similarly across Q will be VAVD
Similarly across R will be VBVC
Similarly across S will be VDVC
By Ohm's law we know that V=IR where I is the current.
VAVB=IP ,VAVD=IQ ,VBVC=IR and VDVC=IS
VB=VD we get
IP=IQ and IR=IS
By dividing the 1st by the 2nd, we get
IPIR=IQIS
PQ=RS this equation is known as the balancing condition in a wheatstone bridge

780649_772566_ans_332d8337044e4e349185fb751a2435a4.png

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