In the circuit shown, the reading of the Ammeter is doubled after the switch is closed. Each resistor has a resistance $ = 1\,\Omega $ and the ideal cell has an $e.m.f. = 10\, V$. Then, the Ammeter has a coil resistance equal to ................ $\Omega$
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When switch is closed reading of ameter is $2 \times 5=10 \mathrm{\,A}$
$\mathrm{R}_{\mathrm{A}}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{10}{10}=1\, \Omega$
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