Question

when a galvanometer is shunted with a 4ohm resistance ,the deflection is reduced to one-fifth .If the galvanometer i further shunted with 2ohm wire,determine current in galvanometer now if initilly current in galvanometer is I'(given main current remains same)

Answer 1

initial current in galvanometer = i

current through the galvanometer after adding shunt of 4 ohm = i/5

so, current through the shunt of 4 ohm = i - i/5 = 4i/5

Let resistance of galvanometer = G

since potential drop across shunt and galvanometer will be same.

hence,

G x i/5 = 4 x 4i/5

=> G = 16 ohm

if the galvanometer is further shunted with 2 ohm resistor, then, net shunt resistance = 4/3 ohm (2 ohm and 4 ohm will be in parallel).

now main current is again "i".
so, "i" will distribute in G and new shunt.

so, let new current through G = X
=> current through shunt = (i - X)

again, P.D. across G and Shunt will be same.

so,

G.X = (i - X)(4/3)

=> 16X = (i - X) (4/3)
=> 12X = i - X
=> 11X = i
=> X = i/11

so, new current through the galvanometer will be (i/11) .