Question

(2)
A galvanometer with a coil of resistance 40 ohm
gives a full scale deflection for a current of 5
mA. How will you convert it into an ammeter of
range 0-5A?​

Answer 1

Answer:

Given: G=400Q

lg=5mA=5x10"3A

I=5A To Find :

=?

Formula :

1-1

Solution :

I-1g 5%107 x 40

5-5x1073

0.04 Q 0.04 Q

A shunt of 0.04 Q is to be connected in parallel with galvanometer (convert if in to am meter of range 0 - 5 A).

Answer 2

Answer:

Galvanometer must be linked in parallel with a 0.04 ohm to convert it in to ammeter of range 0 - 5 A).

Explanation:

Given :

R=40 ohm,

Ig=0.05A and I=5A

FORMULA USED:=

Ig x Rg =(I - Ig) x R

Obtaining results :

According to question ,

0.005 x 40 =(5 - 0.005) xR

=> R = 0.005 X 40/(5 - 0.005)

=> R = 0.04 ohm

Galvanometer must be linked in parallel with a 0.04 ohm to convert it in to ammeter of range( 0 - 5 A).

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