Find the ratio of currents as measured by ammeter in two cases when the key is open and when the key is closed
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When key is open

$\mathrm{R}_{\mathrm{eq}}=\frac{[\mathrm{R}+2 \mathrm{R}][\mathrm{R}+2 \mathrm{R}]}{\mathrm{R}+2 \mathrm{R}+\mathrm{R}+2 \mathrm{R}}=\frac{9 \mathrm{R}^{2}}{6 \mathrm{R}}=\frac{3 \mathrm{R}}{2}$

$\mathrm{V}=\mathrm{IR} \quad \Rightarrow \mathrm{I}_{1}=\frac{2}{3} \frac{\mathrm{V}}{\mathrm{R}}$

When key is closed

$\mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{R} \times 2 \mathrm{R}}{\mathrm{R}+2 \mathrm{R}}+\frac{\mathrm{R} \times 2 \mathrm{R}}{\mathrm{R}+2 \mathrm{R}}=\frac{4 \mathrm{R}}{3}$

$\mathrm{V}=\mathrm{I}_{2} \mathrm{R} \Rightarrow \mathrm{I}_{2}=\frac{3}{4} \frac{\mathrm{V}}{\mathrm{R}} \Rightarrow \frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\frac{8}{9}$

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