A galvanometer shows full-scale deflection for a current $I_g$. With a shunt $S_1$ it becomes an ammeter of range $(0\text{–}I)$; with a shunt $S_2$ its range becomes $(0\text{–}2I)$. Then $\dfrac{S_1}{S_2}$ is:

CBSE Class 12 · 2024 · Current ElectricityMultiple Choice

A galvanometer shows full-scale deflection for a current

I_{g}

. With a shunt

S_{1}

it becomes an ammeter of range

(0 - I)

; with a shunt

S_{2}

its range becomes

(0 - 2 I)

. Then

\frac{S _{1}}{S _{2}}

is:

Answer: (C)

At full-scale the drop across the galvanometer equals the drop across its shunt (they are in parallel):

I_{g} R_{g} = S (I_{line} - I_{g})

S_{1} (I - I_{g}) = I_{g} R_{g} = S_{2} (2 I - I_{g}) \Rightarrow \frac{S _{1}}{S _{2}} = \frac{2 I - I _{g}}{I - I _{g}} .

CBSE Class 12 Board · 2024 · Single correct option