The resistance of the four arms of a wheatstone bridge are `100Omega,10Omega,300Omegaand 30Omega` respectively . A battery of emf 1.5 V and negligible internal resistance is connected to the bridge . Calculate the current flowing through each resistance .

Explanation

Here , `P=100Omega`
`Q=10Omega,R=300Omega,S=30Omega,and E = 1.5 V`

Since , `(100)/(10)=(300)/(30)`
i.e., `(P)/(Q)=(R)/(S)` the bridge is in a balanced condition
`therefore` Current in the resistance P = current in the resistance `(V_(AB))/(P+Q)=(1.5)/(100+10)=(1.5)/(110)=0.0136A`
Again , current in the resistance R = current in the resistance
`(V_(AB))/(R+S)=(1.5)/(300+30)=(1.5)/(330)=0.0045A`.