In the circuit shown, the reading of the ammeter is doubled after the switch is closed. Each resistor has a resistance equal to 1 Ω and the ideal cell has an emf equal to 10 V. Then the ammeter has a resistance equal to

$\begin{array}{l} I_{1} = \frac{10}{2 + R_{A}}, I_{2} = \frac{10}{R_{A}} and I_{2} = 2 I_{1} \\ Solving, we get R_{A} = 2 Ω \end{array}$

In the circuit shown, the reading of the ammeter is doubled after the switch is closed. Each resistor has a resistance equal to $1 Ω$ and the ideal cell has an emf equal to 10 V. Then the ammeter has a resistance equal to