In the given circuit a bridge is shown in the balanced state.The meter bridge wire has a resistance of `1 Omega cm^(-1)` .Calculate the unknown resistance `X` and the current drawn from the battery of negligible internal resistence. If the magnitude the position of galvanometer and the cell, how it will affect the position of the galvanometer ?

Answer

` 0.33A`

Explanation

Here resistance of arm `AB p = 30 xx 1 = 30 Omega`
Resistance of arm `BC = Q = 70 xx 1 = 70 Omega`
when bridge is balance then
`(P)/(Q)= (X)/(Y) or X = (P)/(Q) Y = (30)/(70) xx 7 = 3 Omega `
When bridge is balance ,no current flow through galvanmeter arm .Now the effective resistance of arm `ADC = X = + Y = 3 + 7 = 10 Omega`
Resistance of arm `ABC = 30 + 70 = 100 Omega`
These two arms in parallel , the effective resistance of the circuit is
` R = (10 xx 100)/(10 + 100) = (100)/(11) Omega `
Currect draw from the battery
`I = (3)/(100//11) = 0.33A`
If the balancing position , we interchange the position of galvanometer and the cell , the bridge remains in balance wire .The galvanmeter show no deffective