In an ammeter, $4 \%$ of the main current is. passing through the galvanometer, If shunt resistance is $5…

In an ammeter, $4 \%$ of the main current is. passing through the galvanometer, If shunt resistance is $5 \Omega$, then resistance of galvanometer will be
  1. $60 \Omega$
  2. $120 \Omega$
  3. $240 \Omega$
  4. $480 \Omega$

Solution

Shunt $S=\frac{\mathrm{I}_{\mathrm{g}} \mathrm{G}}{\mathrm{I}-\mathrm{I}_{\mathrm{g}}}$
On substituting the given values, we get, $\begin{aligned} \therefore \quad & 5=\frac{\left(\frac{4}{100} \mathrm{I} \times \mathrm{G}\right)}{\mathrm{I}-\frac{4}{100} \mathrm{I}}=\frac{4 \mathrm{G}}{96} \\ & \Rightarrow \mathrm{G}=\frac{96 \times 5}{4}=120 \Omega \end{aligned}$ .

Asked in: MHT CET 2024 (09 May Shift 1)

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