A galvanometer has a resistance of 50 ohm. A resistance of 5 ohm is connected across its terminals . What part of total current will flow thorugh the galvanometer?

Explanation

Resistance of galvanometer ` = G = 50 Omega`
Shunt ` = S = 5 Omega`
Current flowing through galvanometer ` = I_(g) = ?`
Let I be the total current flowing through the circuit.
Then current flowing through shunt (S) be `(I-I_(g))`
As the shunt resistance is connected in parallel with the galvanometer
`therefore (I-I_(G))S = I_(G)G`
`IS -I_(G)S = I_(G)G`
`IS = I_(G)S + I_(G)G = I_(G) (G +S)`
`I_(G) = (IS)/(G +S) = (I(5))/(50 +5) = (5)/(55)I = (I)/(11)I A`