Question

A galvanometer of resistance $240 \, \Omega$ allows only $4\%$ of the main current after connecting a shunt resistance. The value of the shunt resistance is

• A. $20\, \Omega$
• B. $8\, \Omega$
• C. $5 \,\Omega$
• D. $10\, \Omega$

Answer 1

Answer: (D)

Given, galvanometer resistance $G=240\, \Omega$

Shunt resistance $S=?$
$I_{G}=\frac{4}{100} I$
From figure voltage through the circuit.
$\left(I-I_{G}\right) S=I_{G} G$
or $\,\,\,\left(1-\frac{4 I}{100}\right) S=\frac{4 I}{100} \times 240$
or $\,\,\,\S=\frac{4 \times 240}{96}=10 \Omega$