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

`7.5 m, 8.75m,6.25 m`

Explanation

At balancing
`1.5=kx`
`1.5=IR/Lx`
`1.5=2/30xx3/10xxx`
`x=75 m`

(a) When a resistance of `5 Omega` is placed in services with accumelator balancing length
`1.5=2/((30+5))xx30/10xxximpliesx=8.75 m`
(b) When cell of `1.5 V` is shunted with `5 Omega` resistor.
At balancing length
`5I=2/30xx3/10xxx`
`5xx1.5/((1.5))=7/5 impliesx=6.25m`