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

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(B) Let the resistance of the ammeter be $R_A = x$. Each resistor has resistance $R = 1\,\Omega$.Case $1$: Switch is open.The circuit consists of the

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(B) Let the resistance of the ammeter be $R_A = x$. Each resistor has resistance $R = 1\,\Omega$.Case $1$: Switch is open.The circuit consists of the cell,one resistor in parallel with the series combination of two resistors and the ammeter. The current through the ammeter is $I_1 = \frac{10}{R + (R + R + x)} = \frac{10}{1 + (1 + 1 + x)} = \frac{10}{3 + x}$.Case $2$: Switch is closed.The resistor in parallel with the switch is short-circuited. The circuit now consists of the cell in series with one resistor and the ammeter. The current through the ammeter is $I_2 = \frac{10}{R + x} = \frac{10}{1 + x}$.Given that the reading is doubled: $I_2 = 2I_1$.$\frac{10}{1 + x} = 2 	imes \frac{10}{3 + x}$$3 + x = 2(1 + x)$$3 + x = 2 + 2x$$x = 1\,\Omega$.Thus,the ammeter resistance is $1\,\Omega$.

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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