In the circuit shown,assume the diode to be i | vedclass

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(B) For an ideal diode,it conducts when the potential at the anode is greater than or equal to the potential at the cathode. Here,the cathode is at a

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$20$

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(B) For an ideal diode,it conducts when the potential at the anode is greater than or equal to the potential at the cathode. Here,the cathode is at a constant potential of $3 \ V$.Case $1$: When $V_{i} = 2 \ V$,the potential at the anode $(2 \ V)$ is less than the potential at the cathode $(3 \ V)$. Thus,the diode is reverse-biased and acts as an open circuit. Therefore,the initial current $I_{initial} = 0 \ A$.Case $2$: When $V_{i} = 6 \ V$,the potential at the anode $(6 \ V)$ is greater than the potential at the cathode $(3 \ V)$. Thus,the diode is forward-biased and acts as a short circuit (ideal). The current $I_{final}$ is given by Ohm's law:$I_{final} = \frac{V_{i} - V_{cathode}}{R} = \frac{6 \ V - 3 \ V}{150 \ \Omega} = \frac{3 \ V}{150 \ \Omega} = 0.02 \ A = 20 \ mA$.The change in current is $\Delta I = I_{final} - I_{initial} = 20 \ mA - 0 \ mA = 20 \ mA$.

In the circuit shown,assume the diode to be ideal. When $V_{i}$ increases from $2 \ V$ to $6 \ V$,the change in the current is (in $mA$):

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