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 ............ $\Omega$
Diffcult
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(when switch is opened)
$10 \times 1 =\mathrm{i}_{1} \times(2+\mathrm{x})$
$\mathrm{i}_{1} =\frac{10}{2+\mathrm{x}}$
$\mathrm{i}_{2}=\frac{10}{\mathrm{x}}$
$\therefore \mathrm{i}_{2}=2 \mathrm{i}_{1}$
$\frac{10}{x}=2\left(\frac{10}{2+x}\right)$
$x=2\, \Omega$
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