In the circuit shown below,the current that f | vedclass

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(C) When the switch $S$ is closed,the points $a$ and $b$ are short-circuited. The circuit consists of two loops. Let the potential at $b$ be $0 \ V$.

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(C) When the switch $S$ is closed,the points $a$ and $b$ are short-circuited. The circuit consists of two loops. Let the potential at $b$ be $0 \ V$. Then the potential at $a$ is also $0 \ V$. The $10 \ V$ battery is connected across the $20 \ \Omega$ resistor. The current through the $20 \ \Omega$ resistor is $I_1 = 10 \ V / 20 \ \Omega = 0.5 \ A$ (upwards). The $10 \ V$ battery is also connected across the $10 \ \Omega$ resistor. The current through the $10 \ \Omega$ resistor is $I_2 = 10 \ V / 10 \ \Omega = 1.0 \ A$ (from right to left). At node $a$,the current $I_2$ arrives from the right and splits into two paths: one through the $5 \ \Omega$ resistor and one through the switch $S$ towards $b$. The current through the $5 \ \Omega$ resistor is $I_3 = V_a / 5 \ \Omega = 0 \ V / 5 \ \Omega = 0 \ A$. Therefore,all the current $I_2$ must flow through the switch $S$ from $a$ to $b$. Thus,the current from $a$ to $b$ is $1.0 \ A$.

In the circuit shown below,the current that flows from $a$ to $b$ when the switch $S$ is closed is ............... $A$.

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