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

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(A) Let the resistance of the Ammeter be $x\,\Omega$. Each resistor in the circuit has a resistance of $1\,\Omega$.Case $1$: When the switch is open,t

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(A) Let the resistance of the Ammeter be $x\,\Omega$. Each resistor in the circuit has a resistance of $1\,\Omega$.Case $1$: When the switch is open,the circuit consists of the cell,one resistor in parallel with a series combination of two resistors and the Ammeter. However,based on the circuit diagram,the Ammeter is in series with the right branch. When the switch is open,the current $i_1$ flows through the right branch containing two resistors $(1\,\Omega + 1\,\Omega = 2\,\Omega)$ and the Ammeter $(x\,\Omega)$. The total resistance of this branch is $(2+x)\,\Omega$. Since the cell is ideal and connected in parallel to this branch,the current is $i_1 = \frac{10}{2+x}$.Case $2$: When the switch is closed,the middle resistor is short-circuited or bypassed depending on the configuration. Based on the provided solution image,when the switch is closed,the circuit simplifies such that the Ammeter is in series with only the resistance $x$ and the cell. The current becomes $i_2 = \frac{10}{x}$.Given that the reading is doubled: $i_2 = 2i_1$.Substituting the values: $\frac{10}{x} = 2 \left( \frac{10}{2+x} \right)$.Simplifying: $\frac{1}{x} = \frac{2}{2+x} \implies 2+x = 2x \implies x = 2\,\Omega$.

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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