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 500Ω and cell of internal

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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 500Ω and cell of internal resistance 1Ω . 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

TestHub · Accepted Answer

Here, I = R + r + G + S GS ​ E ​ and I g ​ = G + S I S ​ ​ I g ​ = ( R + r ) + ( G + S ) GS ​ E ​ × ( G + S ) S ​ ∴ I g ​ = ( R + r ) ( G + S ) + GS ES ​ ​ For S = 5 ohm , I g ​ = 5 × 1 0 − 3 A and for S = 25 ohm I g ​ = 20 × 1 0 − 3 A Hence, 5 × 1 0 − 3 = 501 ( G + 5 ) + 5 G E × 5 ​ and 20 × 1 0 − 3 = 501 ( G + 25 ) + 25 G E × 25 ​ Dividing and solving, G = 88.2Ω From (i), we get E ​ = 1 0 − 3 [ 501 ( 88.2 + 5 ) + 5 × 88.2 ] = 47.1 volt ​

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