Question

A galvanometer of resistance 20 ohm is shunted by 2ohm resister .the part of the main current flows through meter is​

Answer 1

Explanation:

Dear buddy

Current= potencial diff

resistance

So resistance=Given resistance + shunt

so here u have to give potencial difference

Answer 2

Explanation:

Given A galvanometer of resistance 20 ohm is shunted by 2 ohm resistor .the part of the main current flows through meter is​

• Given there is a galvanometer and along with it there is a shunt resistor. The main current is i. The current through the galvanometer is ig. So we have I – ig. Since this is in parallel the potential difference of the two will be equal.G is the resistance of galvanometer.
• So we can write as
•               So ig x G = (i – iG) S (S is shunt resistance)
• We need to find the part of main current flowing through the galvanometer that is ig/i
•               So ig x G = iS – iGS
•            Or iG (G + S) = iS
•            Now iG / i = S / G + S
•                              = 2 / 20 + 2
•                             = 2/22
•                             = 1/11
• Therefore iG = 1/11 times of i

Reference link will be

https://brainly.in/question/13264367