Question

20% of the main current passes through the
galvanometer. If the resistance of the galvanometer
is G, then the resistance of the shunt will be -

​

Answer 1

Answer:

S=G/4

Explanation:

S = G.ig/i-ig

S = G * 20/100 - 20

S = 20G/80

S= G/4

this is the ri8 way to solve........

Answer 2

The resistance of the shunt is G/4 ohm.

Given:

Only 20% of the main current passes through the galvanometer.

To Find:

The resistance of the shunt

Solution:

Let S be the shunt resistance and $I_{g}$ is the current passing through the galvanometer.

From the diagram

$V_{AB} = I_{g} G = (I - I_{g}) S$

⇒$S = \frac{ I_{g}G}{(I- I_{g})}$

Ig = 20% I = 0.2I

$S =\frac{0.2I*G}{I - 0.2I} \\\\S =\frac{0.2I}{0.8I} G\\\\S = \frac{1}{4} G$

Therefore, The resistance of the shunt is G/4 ohm.

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