The resistance of a galvanometer coil is R, then the shunt resistance required to convert it into a ammeter of range 4 times, will be

Question

The resistance of a galvanometer coil is R, then the shunt resistance required to convert it into a ammeter of range 4 times, will be (A) 4R (B)  (R/3)  (C)  (R/4)  (D)  (R/5)

Tardigrade Admin · Accepted Answer

Shunt is connected in parallel with galvanometer.
Ammeter is made by connecting a low resistance shunt

S

in parallel with galvanometer

G

. Since

G

and

S

are in parallel, the potential difference across them is same

i_{g} \times G = (i - i_{g}) \times S

Given,

G = R, i = 4 i_{g}

∴ S = \frac{i _{g}}{4 i _{g} - i _{g}} \times R = \frac{i _{g}}{3 i _{g}} \times R = \frac{R}{3}

Note : Resistance of shunt is always low.

Q. The resistance of a galvanometer coil is $R$ , then the shunt resistance required to convert it into a ammeter of range $4$ times, will be

4R

$\frac{R}{3}$

$\frac{R}{4}$

$\frac{R}{5}$

Q. The resistance of a galvanometer coil is R, then the shunt resistance required to convert it into a ammeter of range 4 times, will be

4R

3R​

4R​

5R​

Q. The resistance of a galvanometer coil is $R$ , then the shunt resistance required to convert it into a ammeter of range $4$ times, will be

$\frac{R}{3}$

$\frac{R}{4}$

$\frac{R}{5}$