A battery of emf `epsilon_0=10V` is connected across a `1m` long uniform wire having resistance `10Omega//m.` Two cells of emf `epsilon_1=2V` and `epsilon_2=4V` having internal resistance `1Omega` and `5Omega` respectively are connected as shown in the figure. If a galvanometer shows no deflection at the point `P`, find the distance of point `P` from the point `a`.

Options

A. 0
B. `25 cm`
C. `50 cm`
D. `100 cm`

Answer

d

Explanation

Resistance of wire `AB = 0.02 xx 100 = 2 Omega`

Current in wire `AB, I= (2)/(2+2) = (1)/(2)A`
Let `x` be the resistance of length `AJ` of wire of which galvanometer shown on deflection, then
`V_(A)- V_(J)=x I= x xx 1//2`....(i)
Here, `epsilon_(1) = 2V: epsilon_(2) = 1V: r_(1) = 2Omega : r_(2) = 1Omega`
Current closed circiut having `epsilon_(1)` and `epsilon_(2)` in series ,
`I_(1) = (2+1)/(2+1) = 1A`
pot diff across `C` and `D` i.e.
`V_(C)- V_(D) = epsilon_(1) - I_(1)r_(1) = 2- 1 xx 2 = 0`
Here, `V_(A) - V_(J) = V_(C)-V_(D) or (x)/(2) = 0 or x = 0`
i.e. length of wire `AJ= 0`