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(B) When key $A$ is opened,the branch containing the $4 \ \Omega$ resistor is disconnected from the circuit.Therefore,the current flows only through t

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$2$

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(B) When key $A$ is opened,the branch containing the $4 \ \Omega$ resistor is disconnected from the circuit.Therefore,the current flows only through the branch containing the $5 \ \Omega$ resistor and the ammeter $A$.The total voltage across this branch is $V = 10 \ V$.The resistance in this branch is $R = 5 \ \Omega$.Using Ohm's law,the current $I$ through the ammeter is:$I = \frac{V}{R} = \frac{10 \ V}{5 \ \Omega} = 2 \ A$.Thus,the reading of the ammeter is $2 \ A$.

In the given figure,when key $A$ is opened,the reading of the ammeter $A$ will be ........... $A$.

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