A variable shunt is connected across the galvanometer which has a current sensitivity of 1mA per division and the combination is put in series with a resistance of `500 Omega` and cell of internal resistance `2 Omega`?. For a shunt of 5 ohm, it gives a deflection of 5 division and for shunt of 30 ohm, it gives 25 division. The emf of cell is-----V.

Answer

82.5

Explanation

`I = E/(R + r + (GS)/(G +S))`
and `I_(g) = (IS)/(G +S)`
`rArr I_(g)`
`=E/((R+r) + (GS)/(G +S)) xx S/(G +S)`
` = (ES)/((R+r) (G+s)+ GS)`
For `S = 5 Omega, I_(g) = 5 xx 10^(-3) A`
`rArr 5 xx 10^(-3) = (5E)/((502)(G+5) + 5G)`...... (i)
`rArr 25 xx 10^(-3) = (30E)/((502)(G+30)+ 30G)`..... (ii)
Dividing (i )and (ii), we get
`6[502 (G+5) + 5G] = 5[502 (G+30) + 30G]`
` = 5 xx 502 G + 5 xx 502 xx 30 + 5 xx 30 G`
`rArr 382 G = 5 xx 502 xx 6 xx 4`
`rArr G = 60240/382 = 157.7 Omega`
Now from (i)
`E = (5 xx 10^(-3))/5 [502(157.7 + 5) + 5 xx 157.7]`
`therefore E = [81.68 + 0.79] = 82.47 ~~ 82.5` V