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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
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Verified Answer
The correct answer is:
$10 \Omega$
Given, galvanometer resistance $G=240 \Omega$
Shunt resistance $S=$ ?
$$
\mathrm{I}_{\mathrm{G}}=\frac{4}{100} \mathrm{I}
$$
From figure voltage through the circuit.
$$
\left(\mathrm{I}-\mathrm{I}_{\mathrm{G}}\right) \mathrm{S}=\mathrm{I}_{\mathrm{G}} \mathrm{G}
$$
or $\quad\left(1-\frac{4 \mathrm{I}}{100}\right) \mathrm{S}=\frac{4 \mathrm{I}}{100} \times 240$
or
$$
S=\frac{4 \times 240}{96}=10 \Omega
$$
Shunt resistance $S=$ ?
$$
\mathrm{I}_{\mathrm{G}}=\frac{4}{100} \mathrm{I}
$$
From figure voltage through the circuit.
$$
\left(\mathrm{I}-\mathrm{I}_{\mathrm{G}}\right) \mathrm{S}=\mathrm{I}_{\mathrm{G}} \mathrm{G}
$$
or $\quad\left(1-\frac{4 \mathrm{I}}{100}\right) \mathrm{S}=\frac{4 \mathrm{I}}{100} \times 240$
or
$$
S=\frac{4 \times 240}{96}=10 \Omega
$$
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