Question

When a galvanometer is shunted with a 4 ohm resistance?

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,

$G \times i/5 = 4 \times 4i/5$

$G = 16 \Omega$

If the galvanometer is further shunted with 2 ohm resistor, then, net shunt resistance = $4/3 \Omega$ (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) ."