A $72 \Omega$ galvanometer is shunted by a resistance of $8 \Omega$. The percentage of the total current which passes through the galvanometer is :

JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Current Electricity

72Ω

galvanometer is shunted by a resistance of

8Ω

. The percentage of the total current which passes through the galvanometer is :

Answer: (B)

The shunt formula relates the galvanometer resistance

R_{G}

, shunt resistance

S

, and the current ratio:

S = \frac{R _{G}}{\frac{I}{I _{g}} - 1}

Substituting

R_{G} = 72 Ω

and

S = 8 Ω

8 = \frac{72}{\frac{I}{I _{g}} - 1}

\frac{I}{I _{g}} - 1 = 9

\frac{I}{I _{g}} = 10

\frac{I _{g}}{I} = \frac{1}{10}

Percentage of current through galvanometer:

\frac{I _{g}}{I} \times 100 = 10%

Therefore, the answer is (B).

JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Single correct option