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

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(D) Let the resistance of the Ammeter be $R_A$. When the switch is open,the circuit consists of a $1\,\Omega$ resistor in parallel with a series combi

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(D) Let the resistance of the Ammeter be $R_A$. When the switch is open,the circuit consists of a $1\,\Omega$ resistor in parallel with a series combination of two $1\,\Omega$ resistors and the Ammeter. The equivalent resistance is $R_{eq1} = \frac{1 	imes (2 + R_A)}{1 + (2 + R_A)} = \frac{2 + R_A}{3 + R_A}$. The total current is $I_1 = \frac{V}{R_{eq1}} = \frac{10(3 + R_A)}{2 + R_A}$. The current through the Ammeter branch is $I_{A1} = I_1 	imes \frac{1}{1 + (2 + R_A)} = \frac{10}{3 + R_A}$. When the switch is closed,the resistor in the middle is shorted. The circuit now consists of the cell in series with the Ammeter and the two $1\,\Omega$ resistors in series. The total resistance is $R_{eq2} = 2 + R_A$. The current through the Ammeter is $I_{A2} = \frac{10}{2 + R_A}$. Given $I_{A2} = 2 	imes I_{A1}$,so $\frac{10}{2 + R_A} = 2 	imes \frac{10}{3 + R_A}$. $3 + R_A = 2(2 + R_A) \implies 3 + R_A = 4 + 2R_A \implies R_A = -1\,\Omega$. Since resistance cannot be negative,there is no solution among the given options.

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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