An ideal Zener diode with breakdown voltage of -3 V is reverse biased with a negative input voltage V_i=-5 V. In the circuit shown (C–[Zener]–B–R–A), the magnitude of the voltage difference between points B and A is:

Solution & Explanation

Setup: an ideal Zener with breakdown voltage 3 V is reverse biased by V_i = −5 V. The branch is C–[Zener]–B–R–A, so the input is shared between the Zener (C to B) and the resistor R (B to A). Principle: in reverse breakdown an ideal Zener clamps the voltage across itself at its breakdown value, here |V_Z| = 3 V, no matter how much current flows. The total magnitude across the series combination is the input, |V_i| = 5 V. This 5 V splits between the Zener and R: |V_input| = |V_Zener| + |V_R|. So the magnitude across R, which is the voltage between B and A, is |V_BA| = 5 − 3 = 2 V. This matches option B, 2 V. Trap: the Zener does not drop the full 5 V; it holds only 3 V, leaving the remaining 2 V across the resistor.