1.

A galvanometer has a current sensitivity of 1 mA per division. A variable shunt is connected across the galvanometer and the combination is put in series with a resistance of 500Omega and cell of internal resistance 1Omega. It gives a deflection of 5 division for shunt of 5 ohm and 20 division for shunt of 25 ohm. The emf of cell is

Answer»

47.1 V
57.1 V
67.1 V
77.1 V

Solution :Here, `I=E/(R+r +(GS)/(G+S))` and `I_(g) = (IS)/(G+S)`
`I_(g) = E/((R+r) + (GS)/(G+S)) xx S/(G+S)`
`therefore I_(g) = (ES)/((R+r)(G+S) + GS)`
For `S= 5 "ohm", I_(g) = 5 xx 10^(-3)A`
and for `S=25 "ohm", I_(g) = 20 xx 10^(-3)A`
Hence, `5 xx 10^(-3) = (E xx 5)/(501(G+5) + 5G)`...........(i)
and `20 xx 10^(-3) = (E xx 25)/(501(G+25) + 25G)`.........(ii)
Dividing and solving, G= `88.2 Omega`
From (i), we get
`E=10^(-3)[501 (88.2 + 5) + 5 xx 88.2] = 47.1` volt


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