For the circuits shown in figures I and II,th | vedclass

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(D) In circuit $I$,the switch (key) is open,which means the circuit is incomplete. Therefore,no current flows through the resistor $(I = 0 \ A)$. Acco

Vedclass · Accepted Answer

$0 \ V$ in circuit $I$ and $2 \ V$ in circuit $II$

Vedclass · Answer

(D) In circuit $I$,the switch (key) is open,which means the circuit is incomplete. Therefore,no current flows through the resistor $(I = 0 \ A)$. According to Ohm's law,the voltage drop across the resistor is $V = I 	imes R = 0 	imes 2 = 0 \ V$. Thus,the voltmeter reads $0 \ V$.In circuit $II$,the switch is closed,completing the circuit. The current flows through the resistor. Since the internal resistance of the battery is not mentioned,we assume it to be ideal. The entire electromotive force $(2 \ V)$ appears across the $2 \ \Omega$ resistor. Therefore,the voltmeter reads $2 \ V$.

For the circuits shown in figures $I$ and $II$,the voltmeter reading would be

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