A galvanometer gives full scale deflection with 0.2 A in a coil shown in the figure. The resistance of its coil is 10 Ω . How much shunt resistance is required to convert it into an ammeter to read current upto 1.8 A ?

Question

A galvanometer gives full scale deflection with 0.2 A in a coil shown in the figure. The resistance of its coil is 10 Ω . How much shunt resistance is required to convert it into an ammeter to read current upto 1.8 A ? (A)  1.25 Ω  (B)  2.5 Ω  (C)  5.0 Ω  (D)  10.0 Ω

Tardigrade Admin · Accepted Answer

Potential difference across galvanometer and shunt is same.

For

G

being resistance of galvanometer and

I_{g}

the current across it, the current across

S

is

(I - I_{g})

.
Since,

G

and

S

are in parallel, potential across them is same

I_{g} \times G = (I - I_{g}) \times S

\Rightarrow S = \frac{I _{g} G}{( I - I _{g} )}

Given,

I_{g} = 0.2 A,

G = 10Ω,

I = 1.8 A

\Rightarrow S = \frac{0.2}{1.6} \times 10

= 1.25 Ω

Q. A galvanometer gives full scale deflection with $0.2 A$ in a coil shown in the figure. The resistance of its coil is $10 Ω$ . How much shunt resistance is required to convert it into an ammeter to read current upto $1.8 A$ ?

$1.25 Ω$

$2.5 Ω$

$5.0 Ω$

$10.0 Ω$

Q. A galvanometer gives full scale deflection with 0.2A in a coil shown in the figure. The resistance of its coil is 10Ω . How much shunt resistance is required to convert it into an ammeter to read current upto 1.8A ?

1.25Ω

2.5Ω

5.0Ω

10.0Ω

Q. A galvanometer gives full scale deflection with $0.2 A$ in a coil shown in the figure. The resistance of its coil is $10 Ω$ . How much shunt resistance is required to convert it into an ammeter to read current upto $1.8 A$ ?

$1.25 Ω$

$2.5 Ω$

$5.0 Ω$

$10.0 Ω$