An accumulator of emf `2 V` and negligible internal resistance is connected across a uniform wire of length 10 m and resistance `30 Omega` The appropriate terminals of a cell of emf `1.5 V` and internal resistance `1 Omega` is connected to one end of the wire and the other terminal of the cell is connected through a sensitive galvanometer to a slider on the wire. What is the length of the wire that will be required to produce zero deflection of the galvanometer? How will the balancing length change?
(a) When a coil of resistance `5Omega` is placed in series with the accumulator.
(b) The cell of `1.5 V` is shunted with `5 Omega` resistor?

Answer

A, B

Explanation

potential gradient across wire,
`AB=2/10=0.2V//m`
Now, `V_(AC)=1.5V or (0.2)(ASC)=1.5`
`:. AC=7.5 m`
a. `V_(AB)=(R_(AB)/(R_(AB)+5))xx2=(30/(30+5))x2=12/7V`

`:.` Potential gradient across
`AB=12/70V/m`
Now, `V_(AC)=1.5V`
`:. (12/70)(AC_1)=1.5`
`:. AC_1=8.75m`
b. `V_(AC_2)=V`
or `(0.2)(AC_2)=(5/(5+1))(1.5)`
or `AC_26.25cm`