A galvanometer having a resistance of 8 Ω is shunted by a wire of resistance 2 Ω.1f the total current is 1 A, the part of it passing through the shunt will be

0.25 A

0.8 A

0.2 A

0.5 A

Solution

0.8 A

Potential difference across galvanometer should be equal to potential difference across shunt.

The shunt and galvanometer are connected as shown figure.

Let total current through the parallel combination is i, the current through the galvanometer is i_g and the current through the shunt is ( i - i_g )

The potential difference V_ab ( = V_a- V_b) is the same for both paths, so

i_g G = ( i - i_g ) S

⇒ i_g ( G + S ) = i S

$\Rightarrow \frac{i_{g}}{i} = \frac{S}{S + G}$

The fraction of current passing through shunt:

$\frac{i - i_{g}}{i} = 1 - \frac{i_{g}}{i} = 1 - \frac{S}{S + G} = \frac{G}{S + G} = \frac{8}{2 + 8} = \frac{8}{10}$

The fraction of current passing through shunt

= 0.8 A

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