In the circuit shown,the reading of the Ammet | vedclass

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(A) Let the resistance of the ammeter be $R_A$. When the switch is open,the circuit consists of the cell in parallel with a branch containing one resi

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$2$

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(A) Let the resistance of the ammeter be $R_A$. When the switch is open,the circuit consists of the cell in parallel with a branch containing one resistor $(1\,\Omega)$ and another branch containing two resistors in series $(1\,\Omega + 1\,\Omega = 2\,\Omega)$ along with the ammeter $(R_A)$. The current through the ammeter branch is $I_1 = \frac{10}{2 + R_A}$. When the switch is closed,the resistor in the middle branch is bypassed. The circuit simplifies to the cell in parallel with a branch containing two resistors in series $(1\,\Omega + 1\,\Omega = 2\,\Omega)$ and the ammeter $(R_A)$. The current through the ammeter branch is $I_2 = \frac{10}{R_A}$. Given that the reading is doubled,$I_2 = 2 I_1$. $\frac{10}{R_A} = 2 	imes \frac{10}{2 + R_A}$. $2 + R_A = 2 R_A$. $R_A = 2\,\Omega$.

In the circuit shown,the reading of the Ammeter is doubled after the switch is closed. Each resistor has a resistance $R = 1\,\Omega$ 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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