Question

A battery of emf `E_0=12V` is connected across a `4 m` long uniform wire having resistance `4Omega//m`. The cell of small emfs `epsilon_1=2V` and `epsilon_2=4V` having internal resistance `2Omega` and `6Omega` respectivley are connected as shown in the figure. If galvanometer shows no diflection at the point `N` the distance of points `N` from the point `A` is equal to

A. `5/3m`
B. `4/3m`
C. `3/2m`
D. none of these

Answer 1

Correct Answer - D
Equivalent emf of two batteires `epsilon_1` and `epsilon_2` is `epsilon=(epsilon_1//r_1 +epsilon_2//r_2)/(1//r_1+1//r_2)=((2//2)+(4//6))/((1//2)+(1//6))`
`2.5V`
Now, `V_(AN)=epsilon`
`:. I_(AN)=(R_(AN))=epsilon`
or `((12)/(4+4xx4))(4)(l)=2.5`
Solving this equation we get
`l=2.5/24m`