An ideal Zener diode with a breakdown voltage | vedclass

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(B) In the given circuit, the Zener diode is connected in reverse bias. Since the input voltage $V_i = 5 	ext{ V}$ is greater than the Zener breakdow

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$2$

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(B) In the given circuit, the Zener diode is connected in reverse bias. Since the input voltage $V_i = 5 	ext{ V}$ is greater than the Zener breakdown voltage $V_z = 3 	ext{ V}$, the Zener diode operates in the breakdown region. In this state, the voltage across the Zener diode (between points $C$ and $B$) remains constant at $V_z = 3 	ext{ V}$. The total input voltage $V_i$ is divided between the Zener diode and the resistor $R$. Thus, the voltage across the resistor $R$ (between points $B$ and $A$) is given by $V_{BA} = V_i - V_z$. Substituting the values, we get $V_{BA} = 5 	ext{ V} - 3 	ext{ V} = 2 	ext{ V}$. Therefore, the magnitude of the voltage difference between points $B$ and $A$ is $2 	ext{ V}$.

An ideal Zener diode with a breakdown voltage of $3 \text{ V}$ is reverse-biased with an input voltage $V_i = 5 \text{ V}$. The magnitude of the voltage difference between points $B$ and $A$ is: (in $\text{ V}$)

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